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A TSO PRESENTATION OF A DECOMPOSITION TECHNIQUE 1/1

FOR SOLVING LARGE-SCALE MULTIDIVISIONAL

LINEAR PROGRAMMING PROBLEMS

By

WILLIAM ARTHUR SENTERS ,,

Bachelor of Science

Oklahoma State University

Stillwater, Oklahoma

1975

Submitted to the Faculty of the Graduate College of the Oklahoma State University

in partial fulfillment of the requirements for the Degree of MASTER OF SCIENCE

December, 1978

A TSO PRESENTATION OF A DECOMPOSITION TECHNIQUE

FOR SOLVING LARGE-SCALE MULTIDIVISIONAL

LINEAR PROGRAMMING PROBLEMS

Thesis Approved:

Thesis Advise~r

/;J.F ~~ I

Dean of Graduate College

:1019533

ii

PREFACE

This study was concerned with the development of an interactive

program designed to aid the student in learning the decomposition

technique. The primary objective is to give the student an opportunity

to learn the concepts of decomposition at his own rate and at the time

he chooses. The program allows the student to visualize how a computer

algorithm goes about solving such a problem.

I would like to express my appreciation to my advisors, Dr. George

Hedrick and Dr. Donald Grace for their assistance and encouragement

through the years, and to Dr. Billy Thornton for giving me a solid

start in the field of operations research. Appreciation is also ex

pressed to Dr. Scott Turner for giving his time and encouragement in

being a committee member. I express special gratitude to my wife,

Deborah, for her love and understanding and for being nearby when

needed.

iii

TABLE OF CONTENTS

Chapter Page

I. INTRODUCTION. 1

II. THE REVISED SIMPLEX METHOD. 4

III. THE DECOMPOSITION ALGORITHM 10

IV.

Angular Structure ... Formulation of the Model Development of Algorithm .

DISPLAY DEVICES .

Time-Sharing Option Terminals. IBM 3277 • Decscope . Decwri ter. .

10 11 13

16

16 16 17 18

V. PROGRAM DESCRIPTION 19

Data Set DECOMP. Data Set PAGE. Data Set JESSE . .

VI. SUMMARY AND CONCLUSIONS .

19 20 21

23

BIBLIOGRAPHY 25

APPENDIXES . 26

APPENDIX A - USER'S GUIDE 27

APPENDIX B - INSTRUCTIONS FOR STORING AND CHANGING PROGRAMS 33

APPENDIX C - SAMPLE OF A SHORT SESSION. 36

APPENDIX D - LOGIC BLOCK DIAGRAMS . . . 49

APPENDIX E - LISTING OF THE TUTORIAL TEXT 56

APPENDIX F - LISTING OF THE CONTROL PROGRAM 72

iv

m

1T

NOMENCLATURE

matrix of division k coefficients for corporate constraints

matrix of coefficients for divisional constraints

matrix consisting of the first m columns of B- 1

the (m+j)th column of B- 1

vector of right hand side of division k constraints

vector of relative cost factors of division k

vector of the objective coefficients associated with the

basic variables

the number of corporate or linking constraints

weights on the jth extreme point of division k

simplex multipliers

l t . f th kth d. . . so u lon space or e lVlSlon

jth extreme point of the kth division

set of variables in division k

v

CHAPTER I

INTRODUCTION

In recent years the business world has turned to mathematical

programming for a scientific approach to decision making. This is the

process of representing a particular real life competitive situation in

terms of an operations research mathematical model. The model usually

consists of an objective function of variables which are subject to a

number of functional constraints each representing a limitation of the

organization. These limitations are usually of the form of limits on

production, demands, manpower, machine hours, natural resources, and

also social responsibilities such as standards on pollution and safety.

It is common practice, and will be followed in this report, to refer to

all constraints as constraints on limited resources. When these con-

straints and the objective function can be represented in linear state-

ments, the process is simplified into linear-programming. A simple

technique for linear programming is the Simplex Method. For a small in-

dependent business, mathematical programming can be a simple task of in-

corporating the Simplex Method without resorting to special techniques.

However, in today's world if an organization wants to operate at

an optimum level and expand, it cannot perceive itself as being inde-

pendent from its environment. In other words, it must realize its

organizational and social dependencies. In order for an organization

to operate as a finely tun~d machine it must operate at a level where l

1

2

the limits on its resources are approached but not reached. As a re

sult strict new constraints are introduced into the mathematical model

of the problem. As one can imagine, the model, accurately stated,

could grow in the number of constraints to such a size as to create an

other problem in itself - this problem being that the great number of

constraints makes for inefficient use of computer time and space.

At this point a company has two alternatives. It could reduce

the number of constraints, hence reducing the accuracy of the model,

or it could incorporate one of the many techniques that have been

developed to alleviate this problem and still keep an accurate model

of the situation. Techniques such as generalized upper bounding,

revised simplex, and decomposition provide an effective way of solv

ing a large problem w·ith a special structure with reasonable expendi

ture of computer time and space.

When some or all of the variables can be divided into groups such

that the sum of the variables in each group must not exceed a specified

upper bound, a generalized upper bound technique can be invoked.

A technique developed specifically for use with digital computers

is the Revised Simplex Method, whereby many of the data can be stored

on external devices, making it possible to solve large problems on

small computers.

The scope of this report will center on decomposition which is

a technique for solving multidivisional types of problems. Many texts

and reports have been written on this algorithm, but not enough pro

grams have been written for the use of students to receive hands on

experience. This report is aimed at developing an interactive program

designed to allow the student to study the decomposition principles at

his own level of detail. The student can cover the material quickly

and briefly or request that a detailed explanation be given for a

specific area. He may also review certain areas of trouble or return

to take the session over as often as he wishes. The report develops

a generalized de.composition program that can be used as a tutorial

supplement to a theoretical presentation and give the advanced student

a feel of how the algorithm can be used and interpreted. Hopefully,

it will result in a better understanding and a more efficient use of

the principle of decomposition and linear programming as a whole.

3

CHAPTER II

THE REVISED SIMPLEX METHOD

There are several factors affecting how long the general Simplex

Method will require to solve a linear programming problem. Two of the

most important factors are the number of constraints and the number of

variables in a problem. If n is the number of variables and m the num

ber of constraints, then the maximum number of iterations possible

will be the value of (m + n)!/(m!n!).

The general model for linear programming in matrix form is:

Minimize Z = C • X

subject to: A • X = b

and >= 0

where C is the row vector of the relative cost factors

C = I C 1 , C 2 , . . . , Cn I , X and b are all column vectors such that

xl b 1

x2 b2 X = b =

X b n n

4

and A is the coefficient matrix

A =

a )_ll

. a mn

Recall that the general Simplex Method began with the ent.ire mxn

matrix A. m being the number of constraints and n being the number of

original variables plus slack variables. At each iteration the entire

mxn matrix was updated and stored. Of the n variables only m basic

variables were in the solution.

The Revised Simplex Method was designed to compute only the in-

formation that is currently needed at each iteration and store it in a

more compact form by comprising an mxm basis matrix of the columns

corresponding to these m basic variables.

Let Pj be the jth column of the coefficient matrix A.

The model can now be restated as

n Minimize z = L: c. . x .

j=l J J

n subject to: E P. . X. = b

j=l J J

ar..:i x. >= 0 J

5

With the Revised Simplex Method the updating operation does not

need to be performed on the entire A matrix, which is mxn with n the

number of original variables plus slack variables. Instead, let B be

an mxm matrix comprised of the basic columns Pj.

i.e. B = I pl' P2, ••• , pm

Only B- 1 need be updated by the pivot operations. The Revised

Simplex Method therefore solves for a set of m equations in m unknowns

(basic variables). This set of equations can be denoted by B•X = b B

where XB is the vector of basic variables so that the basic solution

is XB = B- 1 • b.

Giveri the basic matrix B, the linear combination that expresses

any other vector P j is determined by computing the vector Pj = B- 1 •P j

which becomes the jth column of the current iteration. The value of

the objective function for a basic solution can now be written as

Z = CB • Pj = CB • B-1 • Pj. Letting 0 be the vector of objective

coefficients for the slack variables ,CB is the subset of the vector

I C,O I containing the values of the objective coefficients associated

with the basic variables.

To avoid computing P~ for all Z.'s a vector of pricing or simplex J J

multipliers is derived by n = A vector Pj not in the basis

is "priced out" by computing Zj = n • Pj = CB • B- 1 • Pj. Thus, the

Pj can be stored on external devices and brought into core memory only

as needed.

It should be remembered that the vectors C and P. were recorded J

in the original data. The CB vector needed to compute n is a row

yector formed from C. All that is needed to form CB correctly is to

keep track of which variables are in the current basis.

6

At each iteration the only relevant pieces of information are:

l) C~, the vector of cost factors Cj- Zj or Cj - nPj relative to the

current iteration. 2) the elements of the updated column Pj where

Pj = B- 1 • Pj. 3) and the values of the basic variables XB where

XB = B- 1 • b.

Using the above information and formulas, let us derive a summary

of the Revised Simplex Method.

Step 0 - Given:

A - coefficient matrix

b - right hand side

C - coefficients of objective function

Initialize matrix B as the columns associated with the ini-

tial basic variables (usually slack variables requiring B to

7

be initialized as an identity matrix). Form CB and B- 1 as

stated above, compute n as. CB·B-1, compute C~ = Cj-Zj = Cj-nPj.

Step l - Determine the entering basic variable. Find Cs = min

element of C~ where s is the index for the entering variable.

Step 2 - Optimization test

If C >= 0 stop. s

If Cs < 0 compute the updated column P~ = B-1 • P and the s s

new simplex multipliers n = C • B- 1 and the new cost factors B

=

Step 3 - Determine the leaving basic variable.

find r as the

index for the variable being removed from the basis by

finding min b;/a: for a~ > 0. ~ ~s ~s

Step 4 - Update the basic solution. Derive new B-1 and set

X = B-1 • b. B

Return to Step 1.

In Step 4, B- 1 could be derived each time by using a standard

computer routine for inverting a matrix. However, since B and B- 1

change by only one vector from one iteration to the next, it is much

more efficient to derive the new B-1 (denote it by B~~w) from the B-1

at the preceding iteration (denote it by B;ld). To do this, let xk

be the entering basic variable, afk be the coefficient of xk (these

coefficients are determined in Step 2), and r be the index of the

column in the preceding basis that is being replaced. The new B- 1

can now be expressed in matrix notation as B~~w = E ·B;±d where the

matrix E is an elementary matrix, i.e., an identity matrix except that

its rth column is replaced by the vector

n =

nm

E can be written as

E =

1 0

0 1

0 0

0 0

where n. = ~

· -aik/a;k

-a2k/a;k

- afk ' if i # r ,

ark

1 if i = ' r a;k

0

0

0

1

8

Of course, in the actual coding of this method the entire E matrix does

not need to be physically built. Only the n vector need be computed,

which could save considerable storage.

If t is allowed to represent the tth iteration then in general the

inverse of the tth basis can b.e obtained from

B-1 -t -

Until now it was assumed that the matrix B contained no artifi-

cial variables or negative slack variables and was therefore equal to

the identity matrix at the beginning of the procedure.

If "=" and/or ">=" constraints are included in the model, artifi-

cial and negative slack variables must be added as in the regular

Simplex Method. The procedure has to begin with a basis consisting of

an identity matrix that corresponds to either real or artificial vee-

tors. A two-phase approach can 'then be used. If the procedure starts

with artificial vectors, a basic feasible solution must be determined

by Phase I, of which the computation is not included in this report.

Phase I can be interpreted as minimizing the sum of the artificial

variables over the feasible region. If a feasible solution is attain-

able, the artificial variables can be driven to zero.

Once an initial basic feasible solution is found Phase II solves

for optimality by the Revised Simplex Method. If the constraints are

all "<=" Phase I may be bypassed.

CHAPTER III

THE DECOMPOSITION ALGORITHM

Angular Structure

There has been a tremendous increase in the division of labor and

segmentation of management responsibilities in organizations recently.

There is also a tendency for the different divisions of an organization

to become independent of the organization as a whole with their own

goals and restrictions. This lends itself to a special class of prob-

lems called multidivisional, to which most large problems belong ..

Their special feature is that

the problem is almost decomposable into separate.problems, where each division is concerned only with optimizing its own operation. However, some overall coordination is required in order to best divide certain organizational resources among the divisions (4, p.l42).

Decomposition ideas and methods are as old as linear programming (6).

But the first workable decomposition algorithm was introduced by

Dantzig and Wolfe in 1959 (3). The basic algorithm that this report

will refer to is quite simple, at least for those familiar with the

mathematics of linear programming and the Revised Sim~lex Method.

The Decomposition Method can be thought of as having each division

solve its own subproblem and send its proposed solution to a central

coordinator who can coordinate the proposals from all the divisions,

impose the corporate viewpoint, and find the optimal solution for the

10

11

overall organization. This is accomplished, not by explicitly imposing

the corporate constraints on the divisions, but by "economic pressure"

in the form of adjustments to the divisions' profit or cost coeffi-

cients to reflect their use of corporate resources. Therefore, we can

reformulate the model in an angular st~cture as follows:

n Minimize r c •x

k=1 k k

subject to: A1 A2 . A n xo bo

B1 0 . . 0 x1 b1

0 B2 0 x2 b2 <=

. -0 0 • B

n X bn n

where the Aj , j = 1, 2, ... , n are matrices that r·epresent the corporate

(linking) constraints. These constraints link the divisions by making

them share the organizational resources available. And the

Bj' j = 1,2, ••. , n represent the divisional constraints of each divi

sion. (0 are null matrices).

Formulation of the Model

At this point let us assume that the set of feasible solutions for

each division is bounded. The solution space for each division is

bounded by the constraint equations on the divisional resources. These

equations define a "flat" geometrical shape (called a hyperplane) in

n-dimensional space analogous to the line in two-dimensional space

and the plane in three-dimensional space. The simultaneous solution of

two constraint equations defines an extreme point. And since we are

12

restricted to linear models the set of points Xk such that Xk >= 0

and Bk • X k = 1Jk constitute a convex set with a finite number of ex

treme points. Therefore, under the assumption that the set is bounded,

any point in the set can be represented as a convex combination of the

extreme points of the set.

Consider the solution space for kth division; call it Sk.

i.e. Sk = {Xk j-Bk • Xk <= bk and~ >= 0}. Any point in Sk can

be represented as a (convex combination) weighted average of the ex

treme points of Sk. (Let X~= jth e.p. of division k). Then

x* k

= Xj is any feasible point of the kth division, where k

* Therefore, this equation for Xk and the so-called

"normalizing" or "convexity" constraints on the /..~ provide a way of

representing the feasible solutions to division k without using any of

the original constraints. Hence, the overall problem can now be re-

formulated with far fewer constraints as

n n Maximize Z = Z Z

k=l j= l

n n (A ·Xj)

j }: z ·\ = bo

k=l j=l k k subject to:

n /..j z = 1

j=l k

and j

/..k >= 0, k = 1,2, ... ' n

This formulation is completely equivalent to the one given ea.rli-

er. However, since it has fewer constraints, it should be solvable

with much les3 computational effort. It also has as many columns as

13

the solution space S has extreme points, which may be thousands. This

fact does not matter much if the Revised Simplex Method is used, as the

columns to enter the basis are generated only as they are needed.

Development of Algorithm

Recall that with the Revised Simplex Method the vector of simplex

multipliers ~ = CB • B- 1 ) is used in computing the relative cost

coefficients. During decomposition ~ needs to be partitioned as

(;1,~ 0 ) with ;1 associated with the reformulated division constraints

and ~ 0 associated with the convexity constraints. Let m denote the

number of corporate (linking) constraints. Let (B- 1) 1 be the matrix ;m

consisting of the first m columns of B- 1, and let (B- 1 )j be the vector

consisting of the jth column of B-1. Then ~ 1 = C •(B- 1 ) a vector B 1 ;m

and ~ 0 = CB•(B- 1)m+j a scalar.

As in the regular Simplex Method, it must be determined whether or

not the current feasible solution can be improved by pricing out vector

Pj' a vector of A. Vector Pj is priced out as in the Revised Sim~lex

Method by ~ 1 ·Pj- cj.

The usual simplex criterion asks that we find

It should be noticed. that the above equation is independent of

the scalar ~ 0 •

Therefore, the first step at each iteration requires solving n

(number of divisions) linear programming problems of the type that

follows.

Minimize

subject to:

and

(;- •A - cj) X . 1 j j

B •x j j

=

>= 0

14

Step 1- Using the simplex multipliers ~ 1 solve tbe division sub-

problems as above obtaining solutions and optimal objective

values Z .. l

Step 2 - Compute the min Zi + ~o = f j

Stopping rule

If f. >= 0 the optimal solution can now be J

letting xj .th extreme point of division

k = J

calculated.

k and A~ the

weights on these extreme points the optimal solution can

By

be

calculated as ~ (A~·Xk) for every division k where the Xk's .1

are the extreme points of the solution space corresponding

to the A~ in the basis of the corporate problem. This cal-

culation results in a vector for each division, each vector

consisting of the number of elements as there are variables

for that division.

Stop.

Step 4 - If f. < ·0 form the column to enter the basis as J

p.. = j

II

where II is an n component vector with a. one in position

j and zeroes elsewhere and A. is the matrix of coefficients J

of the corporate constraints for division j.

Step 5 - For the Revised Simplex Method to determine the leaving

basic variable it is necessary to calc~late the current

coefficients and right hand side as B- 1·Pj and B- 1·b'.

bo b' being the vector of

1

vector of all l's.

Step 6 - Obtain a new basis inverse.

Obtain new simplex multipliers.

Go back to Step 1 and repeat.

where 1 is an n component

1~

CHAPTER IV

DISPLAY DEVICES

Time-Sharing Option Terminals

Any visual display device that can be used as a time-sharing

option (TSO) terminal can be used to execute this decomposition presen

tation. Most TSO terminals differ only in the way the data is entered

and displayed. Therefore, a basic understanding of the terminal being

used will be helpful, much like one should know how to operate a type

writer before he can learn to type.

Three common devices used with TSO are the IBM 3277, Decwriter,

and Decscope. General·and brief instructions for each are included in

this chapter. There are many models of each and detailed instructions

might differ among them.

IBM 3277

An IBM 3277 is a device that consists of a screen to display out~

put much like a television screen. Instead of displaying one line at a

time, it can display a number of lines at one time. This is referred

to as a pag~. The size of the page may differ with each model but the

most common page is 22 lines long. The user has some control over when

information is displayed. To enter information into the system, the

IBM 3277 utilizes a keyboard. Data entered through the keyboard is

16

17

also displayed on the screen. A cursor indicates where on the screen

information will be displayed. To enter a command or answer a ques

tion, the user types the command or answer on the keyboard and depres

ses the 'ENTER' key. To retype any portion of the line he depresses

the backspace (+) key. However, any mistakes must be corrected before

the 'ENTER' key is depressed.

The program has one peculiarity when 3277 units are used: at

times part of a page will be displayed at one time .and the rest of the

page on the next screen. In order to prevent this from happening, the

'CLEAR' key should be depressed before entering a command. The 'CLEAR'

key will clear the screen and bring the cursor to the top, then the

entire next page can be displayed.

Decscope

A Decscope is similar to an IBM 3277. It too has a keyboard and

screen with a cursor. To enter data into the system via a decscope

the user types the command on the keyboard and depresses the 'RETURN'

key. However instead of displaying a page at a time, the decscope

writes only one line at a time, then spaces it up. As the information

reaches the top of the screen, it is lost. Again, any pertinent infor

mation should be recorded for future reference as it is lost upon

leaving the screen. There is no possibility of only half a page

appearing on the screen at a time; therefore, to continue the session

it is not necessary to clear the screen before displaying the next

page.

18

Decwriter

The Decwriter is a simple typewriter-type terminal with a key

board for input and a. hard copy printer for output. There are various

models of Decwriters varying in the kind of printing mechanism, the

speed of printing, and a number of other aspects. The Decwriter is

similar to the Decscope in that only a line at ~ ~ime is printed. To

enter data the user depresses the carriage returfl after the data are

typed. To learn the details of operating a particular model one

should read the operations manual of that model.

Because the Decwriter uses a mechanical printing device rather

than an electronic display device it is slower than the IBM 3277.

However, it does allow the user to maintain a hard copy of the session

for future reference.

CHAPTER V

PROGRAM DESCRIPTION

The program written in connection with this study is designed to

convey basic ideas about decomposition. This chapter describes the

function of the program, its limitations and some of the problems

encountered. The program was developed to be used on a TSO(time shar

in~ system. Most TSO terminals have a typewriter-like keyboard to

enter data. The features of each keyboard vary from terminal to

terminal.

The program consists of three major TSO data sets working together

to accomplish the desired results. They are named DECOMP, PAGE, and

JESSE.

Data Set DECOMP

A command procedure is a TSO data set of prearranged executable

sequence of commands with a description qualifier of 'CLIST'. The data

set DECUMP is a command procedure or 'CLIST' created to control the

processing of the overall program.

DECOMP is divided into two parts, Part l and Part 2. Part l con

trols the display of the pages of the text. The text begins by giving

the background of decomposition. It follows with a description of the

technique and gradually leads the student through the theory of the

algorithm and an example.

19

20

The program is interactive in that the user can proceed, not only

at his own rate, but to whatever degree of detail he wishes. The pro

gram.is designed to take the student through a general approach to

decomposition. He may request further instruction on any topic, as

needed. The user must read the information and answer questions based

on what he has just learned. The program will immediately tell the

user if he has answered correctly or incorrectly, and either allow him

to proceed or to review the information and attempt to answer the

question again. There are several places where the user can stop and

start over at the beginning if he feels it is necessary or reread pre

vious pages.

At the conclusion of Part l the user has three choices. He can

go through Part l again, terminate the session at that point, or enter

Part 2.

Part 2 lets the user enter his own data to be run through a decom

position program named JESSE. Part 2 may be entered as often as needed

to run more than one-problem.

For greater detail on input and output, consult the User's Guide,

Appendix A.

Data Set PAGE

A 'DATA' type data set contains any unformatted upper case data

of any type. PAGE is a 'DATA' data set that contains all the pages of

the text for Part l of the program. The command procedure DECOMP de

termines when these pages will be displayed. Each p~ge explains ideas

and gives instructions to the user prompting his response to questions.

Briefly, DECOMP controls the interaction betw~en the responpes from

21

the user and the text in PAGE.

The pages begin by indicating the assumptions made about the stu

dent's background in L.P. and Revised Simplex and gives an introduction

to the operations of the program. It then continues with the back

ground of decomposition and a development of the technique.

It then concludes with a step-by-step procedure to solve a decom

position problem and gives an example of the procedure.

Data Set JESSE

JESSE is a data set containing a Fortran program that executes a

decomposition algorithm. It is used exclusively in Part 2 of the over

all program. It allows the student to input the necessary coefficients

to a decomposition problem. As it solves the problem, intermediate

results are printed to allow the student to follow the progress of the

algorithm.

For greater detail on input and output of Part 2, consult the

User's Guide, Appendix A.

Limitations of Part 1

The user has the option of reviewing certain information but the

information to be reviewed is not at the discretion of the user. The

reviewed pages are predefined by the control program. The information

contained in the program is the only information available to the user.

Unlike classroom instruction where some personalized instruction is

available and questions may be asked, programmed instruction limits the

amount of feedback from the student. When using a terminal other than

a Decwriter, the user should take notes to which he can refer later.

22

Limitations of Part 2

The actual program that performs the decomposition algorithm con

tains a few limitations on the type of problem that can be solved. The

problem must have no more than 20 subdivisions and no more than 20

constraints each. It must have no more than 20 corporate constraints

and all constraints must be "less than or equal to" inequalities.

Further Study

The possibilities of refinement of the presentation seem unlimit

ed. A more sophisticated interactive program could be written to

include more questions and examples and even keep a score to judge

the student's progress.

Further study could also be done to incorporate a graphical repre

sentation of the decomposition concepts, as done by Adams (1) for basic

Linear Programming.

Part 2 could be further developed to include problems with "great

er than or equal to" inequalities. Part 2 was written in Fortran which

limits its generality. A program which lent itself to variable dimen

sioning would require fewer limitations on the size of the problem.

CHAPTER VI

SUMMARY AND CONCLUSIONS

This report describes a method whereby the concepts of decomposi

tion can be presented interactively using a time-sharing option (TSO)

terminal. The first chapter is an introduction to the report. It

discusses linear programming and leads into the large scale linear pro

gramming problem. Chapter II reviews the Revised Simplex Method.

Chapter III describes a way of solving a large-scale problem. ·It

covers the decomposition method and the formulation of the decomposi

tion model. It then presents a six step decomposition algorithm.

Chapter IV discusses three common time-sharing terminals that can be

used to execute the program, along with their differences that may

cause some difficulty in operation. Chapter y d~scr~bes the function

and internal operation of the program, its limitations, and some of

the problems encountered. It is assumed that the user has some know

ledge of linear programming, especially of the Revised Simplex Method.

Appendix A is a User's Guide of detailed instructions on the oper

ations of the program. Appendix B gives instructions for storing and

changing the prognam, allowing for changes to the tutorial text. Ap

pendix C is an example of a short session that a user m:ight execute.

Appendix D contains the logic block diagrams of the control data set

and the decomposition program. And Appendixes E and F are listings of

the tutorial text and the control data set.

23

Any organization that has access to a TSO system has access to a

valuable educational tool. A training program can benefit greatly by

using the interactive capabilities of the system for pedagogical pur

poses. With a system of this kind the educational process is not

subject to the inconsistent performance of an instructor. More time

can be spent in preparing the sessions, which may be prepared by many

educators, therefore achieving a more ,efficient presentation. Many

times students contribute greatly to a particular subject during a

24

lab or seminar. These contributions which would otherwise be lost,

can be incorporated in the programmed instructions for the benefit of

future classes. In effect it eliminates the human error factor from

classroom instruction. However, it should be remembered that program

med instruction as discussed in this report is a tool of education

and is not meant to replace classroom instruction. Such a tool is

meant to give supplemental aid to the student, thus allowing the

teacher more time to give individual attention.

BIBLIOGRAPHY

(1) Adams, Marsha Lynn. 11Tutorial and Visual Display of the Basic Concepts of Two Dimensional Linear Programming." (Unpub. M.S. report, Oklahoma State University, 1975.)

(2) Dantzig, George B. Linear Programming and Extensions. Princeton, New Jersey: Princeton University Press, 1963.

(3) Dantzig, George B., and Wolfe, Philip. "Decomposition Principle for Linear Programs." Operations Research. Volume 8, Jan., 1960~

(4) Hillier, Frederick S., and Lieberman, Gerald J. Operations Research. 2nd Edition, Holden-Day Inc., 1974.

(5) Lasdon, Leon S. Optimization Theory for Large Systems. New York: The MacMillan Company, 1970.

(6) Orchard-Hays, William. Advanced Linear Programming Computing Techniques. McGraw-Hill Book Company, 1968.

(7) Wagner, G. R. Learning." 1972), pp.

"Conversational Linear Programming for Experiential Engineering Education. Volume 62, No. 7 (April, 824-826.

(8) Zacharov, B., Davies, B. W., and Chau, A. Y. C. "Island- An Interactive Graphics System for Mathematical Analysis." The Computer Journal. Volume 17, No. 2 (May, 1974), pp. 104-112.

25

APPENDIXES

26

APPENDIX A

USER'S GUIDE

27

DECOMP can be executed on any system containing the TSO time

sharing options. The user must first acquire a TSO user ID from his

computer center and find out the ID number under which the program is

stored. Once this is done he can log on and begin the session in the

following manner.

STEPS

1. To log on any TSO terminal, use the keyboard to key in the

command LOGON aaaaaaa. Where aaaaaaa is the TSO user ID number.

2. Depress the ENTER or RETURN key. (If the number is invalid,

the message INVALID 1PROJECT NUMBER will appear. If this happens, try

Step 1 again. If the situation persists, consult the computer center

about the ID number.

3. If the number is password protected, you will be asked for

the password. Key in the password.

28

4. Depress the ENTER or RETURN key. (If the password is invalid

the message INVALID PASSWORD FOR USER ID aaaaaaa will appear. If this

happens, go to Step 1 again. If it happens again consult the computer

center.)

5. To begin the session, the user should key in EXEC 'TSO.bbbbbbb

DECOMP.CLIST'. Where bbbbbbb is the TSO ID number the program is

stored under at the installation.

6. Depress the ENTER or RETURN key. The progr~ will now begin

executing.

Executing DECOMP

The first screen of information is general instructions on using

the program. Detailed instructions are given.at each step when a

29

response is required by the user.

1. After reading the instructions thoroughly, depress ENTER or

the carriage return. (If an IBM 3277 is used, the program generates a

message tu the user to clear the screen before each enter.)

2. The first page of the text will then appear. It explains that

the session is in two parts, the tutorial text and an executable pro

gram. A choice is given as to which part to execute, enter the appro

priate response. You will then be prompted for your name.

3. Assuming the student wishes to step through the text, the

second page will appear, outlining the main context of the text, along

with additional instructions.

4. Pages 3 and 4 give an introduction into multidivisional pro

blems and explain the angular structure of their constraints. Again

press ENTER or RETURN after each page.

5. Page 5 presents the first question of the session. The mes

sage ANS = will be printed allowing the answer to be entered immedi

ately following the message. Three chances will be given to answer the

question correctly. Upon request the program will return to Page 3 for

a review.

6. Page 6 explains the answer to Question #1.

7. A general description of decomposition is then presented

followed by Question #2. Again three chances are given to enter the

correct answer and an explanation of the answer is given.

8. Page 9 defines a multidivisional problem and the constraints

needed for reformulation. If the student wishes he can view Page 10

for an in depth study of the reformulation.

9. Otherwise he can continue to Page 11 and Question #3 on

30

reformulation. Again the message ANS = is printed to prompt a re

sponse. If all three chances are used he will be given the opportunity

to review Page 10.

10. Page 12 discusses why the simplex multipliers need to be

partitioned for decomposition. Here the student is given the opportun

ity to continue discussion at a more detailed level by requesting to

view Page 13.

11. Or the student can continue to Page 14 and be given Question

#4 on why the simplex multipliers are partitioned. The format for the

response is similar to-Question #2 and #3. And a chance is given to

request a review of Page 13.

12. At this point the student may choose to see the six step

simple algorithm or continue to Part 2. The steps are presented on two

pages and at the end of the second page the student may choose to see

an example of how the algorithm works.

Execution of the Example

The student is take-n through the entire execution of an actual

problem. Intermediate results are given at each step to help the stu

dent visualize the process taken at that point by the use of the inter

active capabilities of the program. Further explanation and actual

computations are available to the student at critical steps.

The first page of the example defines the problem to be solved.

For efficiency purposes a problem was chosen from .Hillier & Lieberman

(4). The problem consists of two divisions of no more than two con

straints each and two variables each. All matrices and vectors are

singled out for clarity. Unless the user is using a decwriter he

should copy this information down for future reference before depress

ing ENTER to continue.

Execution of Part 2

31

Part 2 is a Fortran program that uses the decomposition algorithm

mentioned in Part 1 to solve a problem who's data is entered through

the terminal. All the data to be entered will be asked for by appro

priate prompting messages.

The first bit of information to be entered is a title to the pro

blem. After the title is entered messages will be displayed asking for

the number of divisions and the number of corporate constraints. These

values should be entered as integer numbers without a decimal point.

The right hand side of the corporate constraints will be asked for

next. These will be read with a Fortran format of F5.2, which means

the first value should be entered with a length of no more than 5 di

gits with the decimal point typed and no more than 2 digits to the

right of the decimal point. Insignificant zeroes to the left or right

of the decimal point do not need to be entered. The enter key should

be depressed after each value is typed. This results in entering one

value per line until all values are entered. The rest of the data is

entered in four steps for each division as follows:

1. Prompting message - 'Type # constraints and # variables for

Division 1'

Response - Enter 2 integer values, one per line.

2. Proinpting message - 'Type x coefficients of the objective

function for Division 1'

Response - Enter x number of real values with the decimal

point as described above, one per line.

3. Prompting message - 'Type x coefficients of Division 1 con

straints'

Response- Again enter x number of real values, one per line.

4. Prompting message - 'Type x coefficients of the RHS of Divi

sion 1'

Response- again enter x number of real values, one per line.

32

The preceding steps will be repeated for each division. When all

data are entered a matrix representation of the program will be dis

played giving the user a chance to view the data and then the opportun

ity to reenter the data if necessary.

Once the data are entered correctly the prog·ram solves the problem

using the decomposition method mentioned in Part 1, giving intermediate

results at each iteration.

APPENDIX B

INSTRUCTIONS FOR STORING AND CEJu~GING PROG~ffi

33

34

The sequential data sets that make ~p this program are stored at

the Oklahoma State University TSO library under the 'l'SO user identifi-

cation number of Ul6300A. 'rheir full qualification is as follows:

'TSO.Ul6300A.DECOMP.CLIST' 'TSO.Ul6300A.PAGE.DATA' 'TSO.Ul6300A.JESSE.FORT'

To store the programs under a personal identification number, a

simple copy command on TSO of the form COPY 'TSO.Ul6300A.name.type'

'TSO.aaaaaaa.name.type' is all that is needed (where aaaaaaa is the

personal identification number of the user).

Once the user has stored the data sets a few changes must be made.

At p~esent DECOMP, which controls the flow of the program, lists the

tutorial text by its fully qualified name. The full qualification must

be changed to the user's ID as follows.

STEPS

It is assumed the user has logged on his own TSO ID and copied the

data sets.

1. With TSO in the READY mode, edit the CLIST by the command

E.DECOMP.CLIST.

2. Once in the EDIT mode, enter the following command:

C 10 50000 /Ul6300A/aaaaaaa/ALL (where aaaaaaa is the user's ID number).

This command changes all occurances of a fully qualified data set name

to the user's ID.

3. Get out of the edit mode by entering END S.

Before the programs can be executed, one other change must be

made. DECOMP calls an object module of the fortran program. Assuming

the user has acquired a copy of the fortran source program JESSE.FORT

he must now create an object module as follows.

With TSO in the READY mode, compile the fortran program by

entering the command FORT JESSE. This compile will create an object

module and the program will be ready to execute.

35

If any changes are made to either the CLIST DECOMP or the tutorial

text PAGE, caution must be exercised. There is a close relationship

between these two data sets and a similar relationship must be present

after any changes are made.

Changes may also be made to the source program JESSE, although

this program can be altered as you would any program written in a high

level programming language. Each time the program is altered, a new

object module must be created as above.

APPENDIX C

SAMPLE OF A SHORT SESSION

36

exec deco111p THIS PROGRAM IS DESIGNED TO OPERATE ON ANY TSO TERMINM.. IT IS INTERACTIVE, MEANING THE USER WILL BE PROMPTED FOR A RESPONSE. A DECWRITER IS PREFERRED SINCE YOU CAN MAINTAIN A HARDCOPY OF THE SESSION AND REFER TO IT AT ANY TIME. HOWEVER, DECSCOPES AND IBM 3277'S CAN ALSO BE USE~• THE OPERATION OF A DECSCOPE AND DECWRITER IS SLIGHTLY DIFFERENT THAN A 3277. IF YOU ARE USING A DECSCOPE 0~ DECWRITER, AFTER TYPING A RESPONSE PRESS THE RETURN KEY. HOWEVER, WITH THE 3277 YOU MUST CLEAR THE SCREEN FIRST THEN ENTER YOUR RESPONSE. THE INSTRUCTIONS DURING A SESSION ASSUME YOU ARE USING AN IBM 3277.

IF YOU ARE USING A 3277 OR SIMILAR TERMINAl ENTER CRT

PAGE.DATA A TSO PRESENTATION OF THE

DECOMPOSITION TECHNIQUE OF LINEAR PROGRAMMING

THIS PRESENTATION IS DESIGNED TO GIVE THE ADVANCED STUDENT A BETTER UNDERSTANDING OF DECOMPOSITION. IT IS DIVIDED INTO TWO PARTS.

yes

PART 1. A TUTORIAL TEXT THAT TAKES THE STUDENT THROUGH THE DEVELOPMENT OF DECOMPOSITION. IT IS ASSUMED THE STUDENT HAS A THOROUGH UNDERSTANDING OF LP AND REVISED SIMPLEX.

PART 2. AN EXECUTABLE PROGRAM THAT LETS YOU ENTER YOUR OWN DATA TO BE RUN AND GIVES YOU INTERMEDIATE RESULTS TO ALLOW YOU TO MONITOR ITS PROGRESS.

IF YOU WOULD LIKE TO STEP THROUGH PART 1 ENTE~YES.

IF YOU WANT TO RUN DATA ENTER NO.

TYPE IN YOUR NAME AND HIT ENTER. b i ll

PAGE.DATA DECOMPOSITION

THIS IS A DEVELOPMENT OF THE DECOMPOSITION TECHNIQUE OF LINEAR PROGRAMMING. IT IS ASSUMED THE STUDENT'S BACKGROUND INCLUDES A THOROUGH UNDERSTANDING OF LINEAR PROGRAMMING AND REVISED SIMPLEX. THE TEXT WILL COVER~ .

1. MULTIDIVISIONAL PROBLEMS 2. THEIR ANGULAR STRUCTURE 3. THE DECOMPOSITION APPROACH - THEORY 4. A DECOMPOSITION ALGORITHM 5. AN EXAMPLE

EVERY SO OFTEN A QUESTION WILL BE ASKED OF YOU. TYPE IN THE ANSWER AND PRESS ENTER. IF AT ANYTIME YOU WANT TO TERMINATE PART 1 AN~ GO TO PART 2 TYPE IN STOP AND PRESS ENTER. <PRESS CLEAR AND ENTER TO CONTINUE)

37

PAGE.DATA

DECOMPOSITION I! A TECHNIQUE USED FOR SOLVING PROBLEMS HAVING A SPECIAL STRUCTURE. THESE PROBLEMS ARE CALLED MULTIDIVISIONAL AND THEIR NAME HINTS AT THE TYPE OF STRUCTURE USED, MULTIDIVISIONAL. HENCE, THEY ARE PROBLEHS THAT ENCOMPASS SEVERAL DIVISIONS. THEREFORE, THE PROBLEMS ARE ALMOST DECOMPOSABLE INTO SEPARATE PROBLEHS, WHERE EACH DIVISION IS CONCERNED ONLY WITH OPTIMIZING IT'S OWN OPERATION. HOWEVER, SOHE OVERALL COORDINATION IS REQUIRED IN ORDER TO BEST DIVIDE CERTAIN ORGANIZATIONAL RESOURCES AMONG THE DIVISIONS.

IF YOU WERE TO LOOK AT A TABLE OF CONSTRAINT COEFFICIENTS FOR THIS TYPE OF PROBLEM YOU WOULD FIND THAT THE CONSTRAINTS FOR EACH DIVISION COULD BE GROUPED TOGETHER IN A BLOCK FORHING AN ANGULAR STRUCTURE.

THE NEXT PAGE EXPLAINS THE ANGULAR STRUCTURE OF MULTIDIVISIONAL PROBLEMS AND GIVES AN EXAMPLE. <PRESS CLEAR AND ENTER TO CONTINUE OR TYPE STOP TO TERMINATE>

PAGE.DATA TABLE OF CONSTRAINT COEFFICIENTS FOR MULTIDIVISIONAL PROBLEHS.

I I

~-----~--------1 I I l __ l

I I l __ l

I I I __ ,

CORPORATE CONSTRAINTS ON ORGANIZATIONAL RESOURCES

CONSTRAINTS ON RESOURCES AVAILABLE ONLY TO DIVISION 1

DIVISION 2

LAST DIVISION

EACH SMALLER BLOCK CONTAINS THE COEFFICIENTS OF THE CONSTRAINTS FOR ONE DIVISION. THE LONG BLOCK AT THE TOP CONTAINS THE COEFFICIENTS OF THE CORPORATE CONSTRAINTS FOR THE MASTER PROBLEM <THE PROBLEM OF COORDINATING THE ACTIVITIES OF THE DIVISIONS>.

38

PAGE.DATA

QUESTION ~1 WHAT TYPE OF SPECIAL PROBLEM WAS THE DECOMPOSITION

METHOD DEVELOPED FOR?

ANS =multidivision~l VERY GOOD BILL

PAGE.DATA

THE CORRECT ANSWER IS MULTIDIVISIONAL

THOSE PROBLEMS WHERE THE MAJORITY OF THE CONSTRAINTS CAN BE SEPARATED INTO GROUPS ACCORDING TO THE RESOURCES AVAILABLE.

TO LEARN HOW THE DECOMPOSITION METHOD SOLVES THESE SPECIAL STRUCTURED PROBLEMS PRESS ENTER TO GO TO THE NEXT PAGE.

<OR TYPE STOP TO TERMINATE>

PAGE.DATA

THE BASIC APPROACH IS TO REFORMULATE THE PROBLEM IN A WAY THAT GREATLY REDUCES THE NUMBER OF FUNCTIONAL CONSTRAINTS AND THEN TO APPLY THE REVISED SIMPLEX. THIS VERSION OF THE SIMPLEX METHOD CAN BE THOUGHT OF AS HAVING EACH DIVISION SOLVE ITS OWN SUBPROBLEM AND SENDING ITS PROPOSAL TO THE MASTER PROBLEM.

IF THESE PROPOSALS VIOLATE THE CORPORATE CONSTRAINTS THE DECOMPOSITION TECHNIQUE WILL EVALUATE THAT VIOLATION AND CALCULATE PENALTIES FOR EACH OF THE DIVISIONS IN ORDER TO FORCE THEIR SOLUTIONS TOWARD A CORPORATE OPTIMUM. IN THIS WAY WE CAN COORDINATE THE PROPOSALS FROM ALL THE DIVISIONS TO FIND THE OPTIMAL 'SOLUTION FOR THE OVERALL ORGANIZATION.

PRESS CLEAR AND ENTER FOR QUESTION ~2 OR STOP TO TERMINATE.

39

PAGE.DATA

QUESTION ~2:

YOU ARE IN CHARGE OF BUDGETING A LARGE CORPORATION AND EACH PLANT MANAGER SENDS YOU PROPOSED BUDGET REQUIREMENTS FOR HIS PLANT. BUT, AS IS USUALLY THE CASE, YOU CANNOT MEET ALL THE REQUIREMENTS. AS BUDGETING DIRECTOR YOUR NEXT STEP IS TO:

A. DETERMINE YOURSELF WHAT THE PLANT BUDGETS SHOULD BE. B. CALCULATE SOME KIND OF PENALTY FOR EACH PLANT TO

FORCE THEM TO COME UP WITH AN AGREEABLE PROPOSAL. C. TEAR UP THE PROPOSALS AND HAVE THEM START OVER. D. RUN THE CORPORATE BUDGET AS A WHOLE USING

REVISED SIMPLEX. THEN SEND EACH PLANT ITS BUDGET.

ANS =d SORRY BILL, D IS AN INCORRECT ANSWER. TRY AGAIN, YOU HAVE 2 MORE CHANCES. ANS =b VERY GOOD BILL THE CORRECT ANSWER IS B. YOU WOULD EVALUATE THE VIOLATIONS AND CALCULATE PENALTIES. BUT HOW? HIT ENTER.

PAGE.DATA LET'S DEFINE A PROBLEM WITH N DIVISIONS AS SUCH:

MAXIMIZE

SUBJECT TO:

N SUM <C*X> I=1

A< 1> A<2> ••. A<N> A<N+1)

A<N+2)

X<G> X< 1> X<2>

::r

B<G> B< 1> B<2>

A<2N> X(N) B<N> WHERE THE B,C,X'S ARE VECTORS AND A'S ARE MATRICES. CONSIDER THE SOLUTION SPACE FOR DIVISION K; CALL IT S(K). ANY POINT

IN S<K> CAN BE REPRESENTED AS A WEIGHTED AVG. OF THE EXTREME POINTS OF S<K>.

LET X(J,K> = EP(J) OF DIVISION K AND L(J,K> IT'S WEIGHT. I.E. ANY FEASIBLE POINT X<*,K> = SUM ON J OF <L<J,K>*X<J,K>> FOR

SOME COMBINATION OF THE L(J,K> SUCH THAT e <~ L<J,K) <= 1 AND THE SUM ON J OF ALL L(J,K> IS EQUAL TO 1.

IF YOU WOULD LIKE TO SEE THE PROBLEM REFORMULATED BY USING THESE CONSTRAINTS ENTER YES. yes

40

PAGE.DATA THIS EQUATION FOR X<*,K> AND THE CONSTRAINTS ON THE L<J,K> PROVIDE

A METHOD FOR REPRESENTING THE FEASIBLE SOLUTIONS TO DIVISION K WITHOUT USING ANY OF THE ORIGINAL CONSTRAINTS. HENCE THE OVERALL PROBLEM CAN NOW BE REFORMULATED WITH FAR FEWER CONSTRAINTS AS

N MAXIMIZE SUM SUM L<J,K><C<K>*X<J,K>>

K=1 J

SUBJECT TO: N

SUM SUM L<J,K><A<K>*X<J,K>> K=1 J

AND SUM L<J,K> = 1 J

STUDY THIS REFORMULATION OF THE MASTER PROBLEM FOR AWHILE. THE SYMBOLISM MIGHT BE CONFUSING. THE FIRST SUMMATION <ON K> REFERS TO THE DIVISIONS. THE SECOND SUMMATION <ON J) REFERS TO THE EXTREME

POINTS WITHIN EACH DIVISION.

PAGE.DATA QUESTION ~3 :

ANS =d VERY GOOD BILL

IN THE REFORMULATION OF THE MASTER PROBLEM WHAT DO THE L<J,K>'S STAND FOR?

A. CONSTRAINT COEFFICIENTS B. SIMPLEX MULTIPLIERS C. EXTREME POINTS IN THE SOLUTION D. RESPECTIVE WEIGHTS ON THE EXTREME POINTS

THE CORRECT ANSWER IS D PAGE.DATA

SINCE THIS REFORMULATION HAS FAR FEWER CONSTRAINTS IT SHOULD BE SOLVABLE WITH MUCH LESS COMPUTATIONAL EFFORT. AT FIRST GLANCE IT WOULD SEEM THAT ALL THE EXTREME POINTS <X<J,K>> NEED BE IDENTIFIED. A TEDIOUS TASK TO SAY THE LEAST. FORTUNATELY, IT IS NOT NECESSARY TO DO THIS WHEN USING THE REVISED SIMPLEX METHOD. ALL THAT IS REQUIRED IS THAT THE SIMPLEX MULTIPLIERS <PI> BE PARTITIONED SO THAT YOU CALCULATE ONLY WHAT IS NEEDED.

DO YOU WANT TO LEARN HOW THIS IS DONE IN MORE DETAIL?yes

41

PAGE.DATA RECALL THAT WITH REVISED SIMPLEX THE VECTOR OF SIMPLEX MULTIPLIERS

<PI = CB * BI> IS USED IN COMPUTING THE RELATIVE COST COEFFICIENTS <BI = B INVERSE>. DURING DECOMPOSITION PI NEEDS TO BE PARTITIONED AS <PI1,PIO>. LET NLC DENOTE THE NUMBER OF CORPORATE <LINKING> CONSTRAINTS. LET BI(1;NLC> BE THE MATRIX CONSISTING OF THE FIRST NLC COLUMNS OF BI, AND LET BI<J> BE THE VECTOR CONSI~TING OF THE JTH COLUMN OF BI. THEN PI1 3 CB * BI(f;NLC) AND PI&~ CB * BI<NLC + J).

THE USUAL SIMPLEX CRITERION ASKS THAT WE FIN~ MIN F<J> = <Pit * A<J> - C(J)) X<J> + PI&

THEREFORE, THE FIRST STEP AT EACH ITERATION REQUIRES SOLVING N <NUMBER OF DIVISIONS> LP PROBLEMS OF THE TYPE

MIN <PI1 * A<J> - CCJ>> X<J> +PIG

SUBJECT TO A<N+J) * XCJ> <= B<J>

X<J> )'2 G

PRESS ENTER TO CONTINUE OR TYPE STOP TO TERMINATE. PAGE. DATA

ANS =a

QUESTION ~4 : WHY ARE THE SIMPLEX MULTIPLIERS, PI,

PARTITIONED INTO Pit AND PI& ?

A. TO SAVE COMPUTATIONAL EFFORT B. TO DISTINGUISH BETWEEN THE SIMPLEX MULTIPLIE~S

OF EACH DIVISION. C. TO COMPUTE EACH RELATIVE COST COEFFICIENT D. SO THAT IT IS NOT NECESSARY TO IDENTIFY ALL EXTREME POINTS.

VERY GOOD BILL THE CORRECT ANSWER IS A. B IS AN M*M MATRIX, BUT TO CALCULATE PI1 AND PI& YOU NEED ONLY NLC+1 COLUMNS OF B. WOULD YOU LIKE TO SEE A SIMPLE ALGORITHM AND EXAMPLE?y•s

PAGE.DATA

STEP 1.

STEP 2.

STEP 3.

STEP BY STEP ALGORITHM

USING THE SIMPLEX MULTIPLIERS PI1 SOLVE THE DIVISION SUBPROBLEMS AS ABOVE OBTAINING SOLUTIONS X AND OPTIMAL OBJECTIVE VALUES Z.

COMPUTE MIN Z + PIG = F(J)

STOPPING RULE IF F<J> >= 0 THE OPTIMAL SOLUTIOH IS SUM L<J>•X<J> WHERE THE X<J> 'S ARE THE EXTREME POINTS OF THE SOLUTION SPACE CORRESPONDING TO BASIC L<J>. REMEMBER, L<J>'S ARE THE RESPECTIVE WEIGHTS OH THESE POINTS AND ARE COMPUTED ONLY UPON TERMINATION OF THE PROBLEM BY THE FINAL B INVERSE TIMES THE ORIGINAL RHS.

STOP

<PRESS CLEAR AND ENTER TO CONTINUE)

42

PAGE.DATA STEP 4. IF F<J> < a FORM THE COLUMN TO ENTER THE BASI~ A~

I ACJ>•XCJ) I A' s 1-----------1

I I I

WHERE I IS AN N COMPONENT VECTOR WITH A ONE IN POSITION J AND ZEROES ELSEWHERE.

STEP 5.

STEP 6.

FOR THE REVISED SIMPLEX METHOD TO NOW DETERMINE THE LEAVING BASIC VARIABLE IT IS NECESSARY TO CALCULATE THE CURRENT COEFFICIENTS AND RHS AS BI*A' AND BI*B'. B' BEING THE VECTOR OF I B<0> I

1------1 I 1 I

WHERE 1 IS ANN COMPONENT VECTOR OF ALL 1'S. OBTAIN A NEW BASIS INVERSE. OBTAIN NEW SIMPLEX MULTIPLIERS. GO BACK TO STEP 1 AND REPEAT.

WOULD YOU LIKE TO SEE AN EXAMPLE OF THIS ALGORITHM?yvs F'AGE.DATA

FOR AN EXAMPLE, CONSIDER THIS PROBLEM WITH 2 DIVISIONS

MAXIMIZE Z s. T.

4XC1) + 6XC2) + BYC1) + 5Y<2>

X ( 1 > + 3X<2> + 2Y< 1 > + 4YC2> 2X < 1 > + 3X(2) + 6Y < 1 > + 4YC2>

X< 1 > + XC2> XC 1 ) + 2X<2>

4Y C 1 > + 3Y<2> AND XCJ),Y(J) >= 0

<= <= <= <= <=

AC1 >=I 3 A<2>=1 2 4 A<3>=1 1 1 I A<4>=1 I 2 3 I 6 4 I 1 2 I

c < 1 >=I 4 6 C<2>=1 8 5 B<0>=1 20 I 25

AND X= X<1>,X<2> ANDY Y<1>,Y<2>

COPY THE ABOVE DOWN FOR FUTURE REFERENCE PRESS ENTER.

F'AGE.DATA

8<1 >=I I

2& 25

5 8

12

4 3

5 8<2>=1 8

THE REFORMULATED MASTER PROBLEM REQUIRES ONLY 4 CONSTRAINTS 2 FOR THE CORPORATE CONSTRAINTS AND 1 CONSTRAINT FOR EACH DIVISION THAT REQUIRES THE SUM OF THE WEIGHTS ADD UP TO 1.

<ON A LARGE PROBLEM THIS WOULD BE A SIGNIFICANT SAVINGS>

FOR THE INITIAL BASIC FEASIBLE SOLUTION

1 0 0 0 2G B 0 1 0 0 BI B' = 25 CB ,. <0,e,e,e>

0 0 1 0 0 0 0 1

WHERE B' IS THE RHS OF THE REFORMULATED MASTER PROBLEM. HIT ENTER.

12

43

I

PAGE.DATA

STEP t. USING THE SIMPLEX MULTIPLIERS Pit SOLVE THE DIVISION PROBLEMS REMEMBER PI = CB * BI INITIALLY CB=(G,G,G,G> ~ BI=I=B, SO PI=<&,&,G,G> ~ PI1=<G,G> THE SOLUTIONS ARE : X(1 > = 2 , X<2> = 3 , AND Z(1) -26

Y(1) = 3, Y<2> = G, AND Z<2> = -24 DO YOU WANT TO SEE HOW THE SOLUTION IS COMPUTED?y~s

PAGE.DATA SOLVE DIVISION ~1 :

MIN Z<1> = <PI1 * A<1>- C<1>>X OR MIN <-4,-6>X S.T. S.T.

A(3)X <= B<1>

THE SOLUTION IS X<1>=2, X<2>=3 AND Z<1>=-26

SOLVE DIVISION ~2 :

1 2

X<1> <= 5 X<2> <= 8

MIN Z<2> = <PI1 * A(2) - C<2>>Y s. T.

OR MIN <-B,-S>Y s. T.

A<4>Y <= 8<2> I 4 3 I Y <= 12

THE SOLUTION IS Y<1>=3, Y<2>=G AND Z<2>=-24

F'AGE.DATA

STEP 2. FI~D THE MINIMUM OF Z<N> + PIG = F REMEMBER PI0 = CB * COLUMN<NLC + N> OF BI N BEING THE NUMBER OF THE DIVISION. THEREFORE PI0 DIFFERS ACCORDING TO THE DIVISION.

F = MIN = -26 THEREFORE THE WEIGHTS<PENALTY> ON E.P. <2,3> OF DIVISION 1 ENTERS THE BASIS

DO YOU NEED HELP?yes F'AGE.DATA

SOLVE: Z<1> + PI0 -26 + 0 -26

Z<2> + PI0 -24 + 0 = -24

PAGE.DATA

STEP 3. STOPPING RULE. IF F IS >~ 0 IT IS AN OPTIMAL SOLUTION. STOP.

F = -26 THEREFORE WE MUST CONTINUE. HIT ENTER.

F'AGE.DATA

STEP 4. GENERATE THE COLUMN TO ENTER THE BASIS AS : A'

IF YOU NEED HELP TO GENERATE PAGE.DATA

I A<1 >•E.P.I A' = 1----------1

I I I

HIT ENTER.

THE

3 2 3

COLUMN ENTER YES.y•s

I I 2 I 1 1 1•1 3 I 13 A' 1 1 0 0

I 1 1 I 11 31 I 1 I I 01

44

PAGE.DATA STEP 5. DETERMINE THE LEAVING BASIC VARIABLE. PROCEED IN THE USUAL

WAY TO CALCULATE THE CURRENT COEFFICIENTS AND THE RHS.

BI * A' 11 13

1 e

BI * B' "' 20 25

1 1

THE RATIOS ARE : 2&/11 '25/13' 1/1

THE MINIMUM RATIO IS 1 <THE THIRD ROW>. R = 3. THUS THE NEW VALUES OF CB ARE (0,0,26,0> THE EXTREME POINTS IN THE BASIS ARE : <_,_> (_,_) (2,3) (_,_)

HIT ENTER. F'AGE.DATA

STEP 6. OBTAIN A HEW BASIS INVERSE AND NEW SIMPLEX MULTIPLIERS. I 1 e I

PI1 = CB * BI<1 ;2> = <&,&,26,&> * I G 1 I = <&,<H = PH

THERE ARE MANY WOULD YOU LIKE

PAGE.DATA

1 e e 1 1 e e 1

WAYS TO FIND AN INVERSE. TO SEE AN EASY ONE?y@S

BI' = E * BI WHERE E IS AN IDENTITY MATRIX EXCEPT THAT IT'S KTH COLUMN IS REPLACED BY THE VECTOR H WHERE

-A'<I,K>/A'<R,K>,IF I~=R M = THEREFORE E =

1/A'<R,K>,IF I=R

so 1 0 -11 e BI' = 0 1 -13 e

0 0 1 0 0 0 0 1


*** ITERATION

STEP 1. RESOLVE THE DIVISION PROBLEMS. SOLUTION FOR DIVISION t1 : X(1) 2 X<2> 3 SOLUTION FOR DIVISION t2: Y<1> = 3 Y<2> = G

DO YOU NEED MORE INFORMATION?yes F'AGE.DATA

COMPUTE THE OBJ. COEFFICIENTS FOR DIVISION ( 0 0) * 11 31 - ( 41 6) = ( --4 -6)

12 31

COMPUTE THE OBJ. COEFFICIENTS FOR DIVISION 2 < 0 0 > * 12 41 - < 8, 5 > = < -B -5 >

16 41 HIT ENTER.

1 e -11 e o 1 -13 e 0 0 1 0 e 0 e 1

z ( 1 ) Z<2>

-26 -24

PAGE.DATA

STEP 2. 1-11 I

PIG = CB * BI(3) s <G,G,26,G) * 1-131 = 26 I 1 I

Z<1> + 26 = G I 0 I

101 PI0 = CB * BI<4> 2 <0,0,26,0> * 101 = G

101 Z<2> + 0 = -24 111

F = -24 THERFORE WEIGHTS ON E.P. <3,0) OF DIVISION 2 ENTERS BASIS

STEP 3. STOPPING RULE.

F = -24 DO WE STOP OR CONTINUE?continuv HIT ENTER.

PAGE.DATA

STEP 4. IA(2) * E.P.I

A' = 1-----------1 I I I

STEP 5. 11 0 -11

12 41 131 = 16 41 * 101

0

01 I 61 I

I 61 = 118 I

I 01 I 1 I

61 BI If A' 10 1 -13 01 * 1181 = 1181

10 0 1 01 I 01 I 01 10 0 0 1 I I 1 I I 1 I

THE RATIOS ARE :

9/6 ' 12/18 I 1/G . 1/1 I 1 0 -11 01 1201 I 91

BI * B' IG 1 -13 01 * 1251 1121 . 10 0 1 01 I 1 I I 1 I

10 0 e 1 I I 1 I I 1 I

THE MINIMUM RATIO IS 12/18. R 2. THE EXTREME POINTS IN THE BASIS ARE

F'AGE.DATA

STEP 6: B INVERSE 11 -1/3 -20/3 01

10 1/18 -13/18 01 10 0 1 0t 10 -1/18 13/18 1 I

CB = (0,24,26,0) ; (_,_) (3,0) <2,3)

PI t :a < 0, 4/3 >

WORK THIS YOURSELF TO BE SURE YOU KNOW HOW IT IS DONE. DO YOU WANT TO START OVER AT THE BEGINNING OF THE EXAMPLE? no

(_,_)

46

PAGE. DATA QUESTION ~'5 .•

IF B INVERSE "'

11 -1 /3 10 1/19 10 0 10 -1118

-20/3 01 -13/19 Gl PI1 '" <0,4/3)

1 o I 13/1 B 1 I

11 31 A<O = 12 31

12 41 A<2> = 16 41 C < 1 > = < 4 6) AND C < 2 > = < 8 5 >

COMPUTE THE COEFFICIENTS OF THE OBJECTIVE FUNCTION FOR DIVISION f AND DIVISION 2 RESPECTIVELY.

A. <O,O> AND <0,0) B. <-4,-6) AND <-8,-5) C. (-4/3,-2) AND (0,1/3) D. <-16/3,-1 > AND <4,-2/3)

ANS =c VERY GOOD BILL THE CORRECT ANSWER IS C. Pit * A - C


*** ITERATION 2

STEP 1. DIVISION OBJ. COEFFICIENTS

(0,4/3) * 11 31 - (4,6) 12 31

(-4/3,-2)

SOLUTION X<t > = 2 X<2> 3 z ( 1 ) -2613

DIVISION 2 OBJ. COEFFICIENTS <0,413> * 12 41 - <a,s>

16 41 (0,1/3)

SOLUTION Y<1> = 0 Y<2> = 0 Z<2>

HIT ENTER. PAGE.DATA

STEP 2. 1-20/3 I

PIO = <0,24,26,0) * 1-13/191 = 26/3 I 1 I I 13/191

Z<1> + 26/3 0 101

PIO = (0,24,26,0) * IOI 0 101 I 1 I

Z<2> + 0 0

F = 0

STEP 3. STOPPING RULE •

F IS >= 0 THIS IS AN OPTIMAL SOLUTION. HIT ENTER.

47

PAGE.DATA THE EXTREME POINTS IN THE BASIS ARE <_,_) (J,G) <2,3) (_,_}

THE WEIGHTS ON THESE POINTS ARE : 5 , 2/3 , t , t/3 COMPUTED BY B INVERSE * ORIGINAL RHS

X SUM L < J) * X< J > = 1 * < 2, 3) <2,3) = X<1>,X<2)

y SUM L< J> * Y( J> 2/3 * <3,0) = <2,0) = Y<1>,Y<2)

THUS, AN OPTIMAL SOLUTION FOR THIS PROBLEM I3 X<1> =2, X(2) =3 Y< 1 > 2 , Y<2> = &

HIT ENTER. IF YOU HAVE DATA YOU WANT TO RUN AS A PROGRAM ENTER YES. no TO END THE SESSION ENTER LOGOFF READY

48

APPENDIX D

LOGIC BLOCK DIAGFJ1.MS

DECOMP

WRITE EXECUTE INSTRUCTIONS

READ REPLY

WRITE INSTRUCTION

READ REPLY

WRITE PROMPT READ

USER'S NAME

WRITE INVALID ANSWER

OUTLINE TEXT

DESCRIBE THE DECOMPOSITION

TECHNI UE

NO

READ REPLY

50

EXPLAIN ANGULAR STRUCTURE OF

CONSTRAINTS

WRITE QUEST #l READ ANSWER

SET CNT=2

DISCUSS ANS. WRITE PROMPT

SUB l FROM CN READ ANSWER

SUB l FROM CNT READ ANSWER

51

WRITE QUEST #2 READ ANSWER

SET CNT=2

DISCUSS ANSWER WRITE PROMPT

READ REPLY

PRINT DEFINITION OF PROBLEM

WRI'rE PROMPT READ REPLY

#3

SET CNT=2

DISCUSS ANSWE WRITE PROMPT

READ REPLY

DETAIL THE REFORMULATION

WRITE MESSAG SUB l FROM CNT READ ANSWER

DETAIL THE REFORMULATION

DETAIL THE PARTITIONING

PARTITIONING OF SIMPLEX MULTIPLIERS

WRITE PROMPT READ REPLY

F

52

WRITE QUEST #4 READ ANSWER

SET CNT=2

ISCUSS ANSWER WRITE PROMPT

READ REPLY

PRINT EXAMPLE

REFORMULATE PROBLEM

WRITE MESSAGE SUB l FROM CNT

READ ANSWER

SHOW HOW TO SOLVE SUBDIV

PROBLEMS

SHOW HOW TO COMPUTE

MIN Z+PIO+F(J)

SHOW HOW TO GENERATE

COLUMN TO ENTER

PRINT STOPPING RULE

PRINT THE COLUMN TO

ENTER THE BASI

53

DETERMINE LEAVING BASIC

PRINT NEW

PRINT NEW SOLUTIONS TO SUBDIVISIONS

COMPUTE MIN Z+PIO+F(J)

SHOW HOW TO FIND INVERSE

SHOW HOW TO SOLVE SUBDIV

PROBLEMS

WRITE MESSAG FOR

WRONG ANSWER

WRITE MESSAGE UB 1 FROM CNT READ ANSWER

PRINT COLUMN TO ENTER THE BASIS

DETERMINE LEAVING

ASIC VARIABLE

COMPUTE NEW INVERSE AND SIMP. MULT.

SET CNT=2

54

DISCUSS ANSWER WRITE PROMPT

READ REPLY

COMPUTE MIN Z+PIO+F(J

DETERMINE THAT THIS IS OPTIMAL

OMPUTE ANSWER

STOP TO END SESSION ENTER LOGOFF

YES

YES

ENTER TITLE TO PROGRAM

55

FREE FILES LOCATE FILES

CALL JESSE A.i'IJD

R PROGRAM

TO END SESSION ENTER LOGOFF

APPENDIX E

LISTING OF THE TUTORIAL TEXT

56

PAGE 1

A TSC PkESENTATICN OF THE DE~vMPuSITiu• TtChNIQvE

CF LL\EAR PRCGRAMI'!It-IG

57

THIS P~ESE~TATION IS D~SIG~tJ TC ~IVE THE ADVANCED STUDENT A EETTER UNUEKSTA~CING 0F DECG~POSITiuN. IT IS CIVIDED INTO T•O PARTS.

PAKT l. A TUTUKIAL TEXT THAT TAKES THE STuDENT THROUGH THE DEVELLFME~T GF DECUMPOSITION. IT IS ASSUMED THE STUDENT HAS A THUKOUGH UNuEKSTANUINb GF LP AND ~EVISED SJMPLEX.

PArtT 2. AN EXECUTAolE PRC~RAM THAT LETS YLU ENTER YCUR u~N DATA TC dE ~UN AND GIVES YUU INTEMMEDIATE MESULTS TO ALLUW YLU TU MuNlTUk ITS PkUGRESS.

IF YwU WUULD LIKE TO STEP THMOUGH PA~T 1 ENTER YES.

IF YCU wANT TG RUN DATA ENTEK NO.

PAGE .::.

t...ELCMPLSITIJN

ThiS IS A CEVELGPMENT LF THE DECuMPCSITIUN TECiiNiQUE OF L&~EA~ PROGkAM~IhG. IT IS ASSUMED THE STUCENT 1 S ~ALKGRUUND INCLUDES A THLkuuGH UNDERSTANDING UF LlN~AR PRUGrtAMMING AND kEVISED SIMPLEX. ThE TEXT wiLL CGViR;

1. MULTIDIVISIC~Al PROtiLEMS L. THEIR ANGULAR STkUCTUkE 3. THE OECLJMPO SIT I c.~ APPkJAC.H - THEURY 4. A utCGI-\PCSI TICN Ali>LRI fh/~ ~. AN t.XAMPLE

EVtkY SO uFTtN A ~UcSTIGN WILL dE ASKED GF YCU. TYPE IN THE ANSWER A.\JC PKtSS ENTtR. If AT ~NYTlMc YL~ NANT Tu TERMINATt PART 1 A~~ ~0 TO PART 2 TYPE If'.; STwP 1-1ND PHESS ENTER. IP~ESS CLEAR AND :NrER TG CCNTINUE)

PAGE 3

DECOMPOSITION IS A TECHNl~E uSED FG~ SGLVlNG PROBLEMS HAVING A SPeCIAL STRUCTURE. THESE PRCdLt~S ARE CALLeD MlLTIOIVlSlONAl ANO THEIR NA~E HI~TS AT THE TYPE Jf ~TRUCTURE USEO, ~ULTlDIVlSiuNAL. HENCE, THEY ARE PROBLEMS THAT ENCLMPASS SeVERAl GIVISICNS. THEREfUR£, THE PKUULEMS ARE ALMOST DECOMPUSA8Lc INTO SEPARATE PROBLEMS, wHERE EACH J I VIS I ON IS CCo\~E RhELl CM. Y 1i ITH U H HH.l.lfliG I Jl S Clllii~ OPERAfi OH. hOWEVtR, SGME CVERALL CCCRDI~ATIUN IS REQUIRED IN ORDER TO oESI DIVIGE C2RTAI,.., ORGANIZATIONAL RE~GURCES AMONG THE DIVISIONS.

IF YOU wERE TO LOOK AT A TABLE OF CCNSTRAihT COEFFICIENTS FOR THIS TYPE OF PROolEM YOU WOULD FIND THAT THE C(;NSTRAlNTS FOR EACH DIVISION COULD BE 6ROUPED TCGEThER IN A JLOCK FCRMING AN ANGULAR STRUCTURE.

THE NEXT PAGE EXPLAINS THE ANGULAR STRUCTURE OF MULTIDIWISlUNAL PROoLL~S AND biVES AN EXAMPLE. (~RESS CLEAR AND ENTER TC CONTINUE OR TYPE STGP TO TERMINATE)

PAGE 4

TAjlf UF CLNST~AlNT COEFFICIENTS FOR MULTIOIVISIONAL PROBLEMS.

I I l_l

I I l_l.

I I I_ I

CURPORATE CCNSTRAI,..,TS CN JRGANIZATlONAL RESCURCES

CONSTRAINTS lN RESCURCES AVAILABLE ONLY t.HV IS ION 1

DlVISIGN 2

LAST DIVISION

EACH SMALLER BLCCK CG~TAINS THE COEFFICIENTS OF THE CONSTRAINTS FOR Cl'.t: DIV-ISiuN. THE ll]lljG 2UlCK AT THE TOP CONTAINS TI-lE CUEffiClcNTS Uf THe CURPORATE CCt.STI'.dii.TS fU.{ THE KASTEl\ PRuBLEM lTHE PRGBLEM Of CClJKOINATlNG ThE ACTIVITIES Gf lHE tHVlSIONSJ.

PAGE 5

~UESTION •1 :

PA~E 6

wHAT TYPE uf SPECIAL PRCBLEH WAS THE UECGMPGSIT!UN METHOD DEVELOPED FOR?

THE CLRRECT A~SWER IS.MULTIDIVISIONAL

THUSE ?RLBLEMS ~HtRE THE MAJORITY uf THE CONSTRAINTS CAN BE SEPA~ATED INTO GROUPS ACCOKUING TO THE RESCUKC2S AVAILA~LE.

TO LEARN HCW THE DECCMPOSITIDN METHOD SOLVES THESE SPECIAL STRUCTUKED PRG6LtMS PRESS ENTER TO GO TO THE NEXT PAGE.

(OR TYPE STCP TO TERMINATE)

59

PAGE 7

THE 3AS1C APPRGACH IS TO nEFuRHULATE THE PRG6LEM IN A WAY THAT GREATLY KEDUCES THE iruMd~K U~ ~UNCTIONAL CCNSTRAINTS ANU THEN TO APPLY THE HEVISt:O SIMPLEX. THIS VERSIGN Uf THE SI/<IPLt:X METHOD CAN Lt THuU~HT Gf AS HAVI~G E~CH DIVISl~N SCLVt iTS CWH SUBPRL~LEH AND SENDING ITS PRUPGSAL TO THE MASTER PRUBLEH.

60

IF THESE PRUPLSALS VlulATE THE CORPCRATE CCNSTRAINTS ThE CS:COMPOSITIUN TECHiH.JUE nlLL eVALUATE TrlAT VIOLATION AND CALCULATE PtfMLTIES FGR EACH Of THE Dl'IISILINS IN GRuER TC Fllri.I.E ThEIR SOLUTIONS TL~AKO A CURPC~ATt GPTIMU~. IN THIS WAY WE CAN CCORDINAT~ THE PRUPLSALS FROM ~ll THE ClVISIU~S TO FIND THE JPTiMAL SOLUTION FOR THE CVEKALL QRGANIZAflUN.

PkESS CLEAR A~O cN7Erl. f~rl. ~U~STION N2 C~ STOP TC TER~lN~TE.

PA~E 8

'-Ut:ST lLo~ 1#2.:

YUU ARE IN CHAKG~ UF JUO~ETING A LARGE CCRPORATION AND c~CH PL~~T MANA~iR SENDS YOU PRUPUSED dUDGEf RE~0l~EMENTS FUK hiS PLA~T. BUT, AS IS USUALLY THt CASto Y0U CA~NUT AE~T All THE ~EQUIRE~tNTS. AS tlUll&ETI~~ ul~E~TuR YJUR NEXT STEP IS TO:

A. DET~q~INE YGU~SELF ~rlAT THE PLA~T BUDGETS SHOULU Sf. a. ~ALCULAfc 5UMc KINO JF PENALfY FOR EACH PLANT TC

fUkCE THEM TG COME UP WITH AN AGREfABLE PROPOSAL. C. TEAR UP THE P~OPOSALS AND HAVE THEM START OVER. O. ~UN THt CGR~GKATE ~UDGET AS A WHULE USING

R2Vl5tD SIMPLEX. THEN S2ND EACH PLANT ITS BUDGET.

PAGE o.;

LET 1 S DEFINE A PRQdlE~ wiTH N DIVISIONS AS SUCH:

MAXI Ml lE

SUtlJECT TO:

N SUM (C(li*Xlll) 1= 1

A(1) A(2) ••• A(N) A( N+ 11

A(N+21

X(O) XU) X(2)

=

8(0) BUJ IH 2)

. . A(2N) X(N) B(N)

WHeRE THE 8,C,X 1 S ARE VECTORS AND A'S ARE MATRICES.

61

CG~SIDER THE SOLuTION SPACE FOR DIVISIGN K; CALL IT S(K). ANY POINT IN SIKJ CAN 8E REPRESENTED AS A MtiGHTED AVG. Cf THE EXTRE~E POINTS Cf SlKJ.

LET X(J,KI = EP(JJ OF !JIVISIJN K AND UJ,KJ IT'S WEIGHT. I.E. ANY FEASlciLE POINT X(*,K) = SUM Oh J OF (L(J,Kl*X(J,KJJ FOR

s,JME ~lJMoiNATIOI\ CF THE LIJ,KJ SUCH TI-AT 0 <= LlJ,KJ <= 1 AND THE SUM GN J GF ALL L(J,KJ IS EQUAL TO 1.

t>AGE 10

THIS =~UATICN FCR X(* 1 KJ AND THE CCNSTRAINTS GN THE L(J,KJ PROVIDE A METrlUu FOR ktPRESENTING THE FEASIBLE SCLUTI(~S TO DIVl~ION K wiTHOUT USli~G Al\Y GF ThE ORIGINAL CUNSTRAINrS. hENCE THE OVERALL PRuBLEM LA~ N~W BE REFORMULATED WITH FAR FEWER CChSTRAINTS AS

N r1AX1 MIZE SUM SUM L(J,KJ(ClKI*X(J,KIJ

K=l J

SUi:iJEC T Tu: N

SUM SUM L(J,KJ(A(Kl*X(J,K)) K=l J

A:'-ll.l SUM LIJ ,K l :: 1 J

STUuY THIS ~EfuKMULATIGN OF rHE MASTER PRGBLE~ fOR AwHILE. THE ~YMBOLISM MIGhT oE CCII.FUSING. THE FIRST SUMMATIGN to/li Kl REFERS TO THE DIVISiuNS. THE SECONil SUMMATION (C." J) REHRS TO ThE EXTR~E PLINTS wiTH!~ EACH DIVISIGN.

PAGE ll

PAGE 12

~UESTIC~ t3 : IN THE REFCRMULATION Of THE MASTER PRGbLEM r~AT OG THE LCJ,A)'S STA~~ fGR 1

A. ~CNSTRAINT CGEFFICIENTS c. SIMPlEX MUlTIPliERS C. EXTREME POINTS IN THE SGLUTICN D. ~ESPECTIVE ~EIGHTS ON THE EXTREME PGINTS

SINCE THIS RtfJRMULATION HAS rAR FEWE~ CGNSTRAlNTS IT SHOUlD 8~ SuLVAbLE ftiTH MUCH LESS COMPUTATIO-Al EffGRT. AT FIRST GLANCE IT ~CUL~ SEEM THAT ALL THE tXTREHE POINTS CXCJ,K)) NEEO dE IDENTirltO. A TEDIOUS TASK TO SAY THE LEAST. fGKTUNATELY, IT 15 NUT NECESSARY TG DO THIS ~HEN USI~G THE REVISED siMPLEX

62

McTHuO. All THAT IS RE~lRED IS THAT THE SlMfLEX HUlTIPLIE~S CPI) ~E

PARTITIONED SO THAT YGU CALCULATE uNLY WHAT IS ~EEDEO.

PAGE ll

PAGE 12




62



PAGE ll

PAGE 12




62



65

PAGE 17

FOR AN EXA~PLE, CGNSIDER T~lS P~CBLEH WITH l DIVISIONS

MAXIMIZE l = 4X(ll + 6X(L) + 8Yl1l + 5Yt2l S. T.

X( 1) + 3X( 2) + 2YIU + 4YI2J 2X( lJ + 3X( 2 l + 6Y ( 1) + 4Yl2J Xll) + XC2) Xl lJ + 2Xl2J

4YUJ + 3'1(2) AND X(J),Y(J) >=O

<= 20 <= 25 <= 5 <= 8 <= 12

A( U=l 1 3 At2l=l 2 4 Al3J=I l 1 I Al4l=l 4 3 I 2 3 I 6 4 I 1 2 I-

Ct ll=l 4 6 CC2l=l il 5 tHuJ=I 20 I E! Ill= I 5 IH2J= I I 25 I I il

ANU X= Xll),X(2) ANDY= Y(1),Y(2)

CJPY THE AbOVE ~L~h ~~R FUTURt REfERENCE

PA<.ic 18

FUR

tl

"HtRE

THE REFLJRMLLATEIJ MASTER PROBLEI'4 REQUII<ES CNL Y It CONSTRAINTS l FGK THE LCrtPuRATE CONSTRAINTS AND 1 CO~~TKAINT FGR EACH ulVISICN THAT ~EQUlRES THE SUM Of ThE WEIGhTS ADO UP TO 1.

t u,~ A LARGE Pi'!UBLEM THIS WUULD BE A SIG~lf I CANT .SAVINGS)

Trh: H .. ITIAL BASIC FEASIBLE SOLUTIG~

l 0 0 0 20 = 0 1 (j 0 = IH B' 25 CB .. co,o,o,oJ

0 0 1 0 l a 0 0 1 1

d' IS T tiE RHS Uf THE REFURMULAT ED HASTER PRGI::ILEM.

12 I

66

PAGE 19

STtP 1. USI~G The SIMPLEX MULTIPLIERS Pll SOLVE lHE DIVISION PROdl~HS ~cMEMBER PI = CB * dl l~lTIALLY Cd=(O,O,O,Ol & ~I=l=B, SO Pl=lO,O,OtOl & Pll=lO,OJ THE SOLUTIC~S Akc : Xlll = 2 , X(~) = 3 , ANU l(l) = -26

Y(l) = 3 , Yl2J = 0 , AND Z(ZJ = -24 SCLVE DIVISIGN Ml :

MIN lll) = lPil * All) - Cl1))X I UR MIN (-4,-b)X S. T. I s.T.

AI3)X <= Blll I 1 1 XllJ I. l 2 X(2J

THE SOLUTICN IS Xlll=2t X(2)=3 ANO ZUJ=-26

SOLVE OIVISIGN M2 : MIN liZ) IPll * A(2) - C(Z))Y OP. MIN (-8,-5JY S. T. s.r.

<= <=

A(4)Y (= 8(2) 1 4 3 I Y <= 12

THE SULUTICN IS Yll)=3, YI2J=O A~D l(2)=-24

t'AGE ZJ

STEP 2. FIND T~E MINIMUM Of llN) + Plu = F kEMEMBER PIO = CB * CuLUMNlNLC + N) Cf Bl N ~tlNG THE NU~BER Of THE DlVlSICN. THEMEF~RE PlO UIF~ERS ACCORDING TO T~E OlVlSlGN.

F = ,"'lN = -26 TrlEkEFORE THt wEI\iHTS(PENALTYJ ON t.P. 12,3) u~ UlvlSION 1 ENTERS ThE eASlS

SOLVE: llU + PIO -26 + () -2b

zc 21 + Piu -24 + I) = -24

!)

~

STEP 3. STOPPING RULE. IF F IS >= 0 IT IS AN GPTIMAL SOLUTION. STOP.

F -26 THEREFORE wE ~UST CONTINUE.

67

PAGE 21

1111 STEP 4. GENEf(ATE THE CCLUHN TO ENTER HiE dASlS AS : A' ::0: U31

I 11 I Ol

I A( U*f:.P. I 1 3 I I 2 I 11 A' = 1--------1 2 3 I* I 3 1 13 = A'

I 1 I 1 1 0 c

SHP 5. DETERMINE THE lEAVING dASIC ~ARlABLE •. PROCEED IN THE USUAL WAY TO ~AlCULATE THE CURHE~T COEfflCIE~TS AND ThE RHS.

B I * A' 11 13

1 0

IH * ~· 20 25

1 1

THE RATILS Ai<E : 2C/11 , 25/13 t 1/1

THE ·"'INIMUM RATIC IS 1 (THE THIRO I{UIIt). R = 3. THUS THE NEk VALLES CF CB ARt (0,0,26,01 THE EXH.tME PuiNTS 1.'1 THf: bASIS ARI: : l_,_) &_,_I (2,3) &_,_)

PAGE 22

STEP 6.. utiTAIN A r>E:W BASIS INVERSE AND NEW SIJIIPLEX MULTIPLIERS. I 1 o I

Pll =Co* ol(l;21 = (0,0,26,01 * I 0 1 I= (0,0) = Pl1 I o o I I o o I

Ul' = E * ai WHfKE E IS AN IDENTITY MATRIX EXCEPT JnAT 1T 1 S KTH CLLJ1'1N 1 S REPLAl.ELI oY THE VECTOR 14 kHERE

1 u -11 0 I -,..• (I oKI/A' lR,KJ ,IF J-,=R 0 1 -13 0 I

M = THEREFURE t: = 0 0 1 0 I 1/A'(R,K),IF I=R 0 0 0 1 I

SLJ 1 I) -ll 0 dl' = I) 1 -13 v

0 0 l 0 0 \) 0 1

i>AGE 23

*** ITERATION 1

STEP 1. R~SULVE THE DIVISION PRuBLEMS. SULuTIO~ FGR DlVl~luN *i : ~ll) = 2 SGluTlGN FOR DIVISION M2 : YCll = 3

XC2J = 3 Y(2j 0

COMPUTE THE CoJ. ~LtfFlLlE~TS FOR DIVISIGN 1 CO 0) * 11 31- C4,6l = C-4 -6)

12 31

" Cu~PUTE THE GdJ. CCEFFICIENTS FGR DIVISION 2 (0 u) * 12 'tl - Cb,5) =·C-8 -.H

16 41

PAl>E 24

STEP 2. 1-111

p 10 = CB * BIC 5) = co,c,26,oJ * l-131 = 26 I 1 I

l ( l) + 26 0 I 0 I

I o I PlO = CB • dl( 4) = co,o,zo,oJ • IOI = 0

IOI lC2) + 0 = -24 I l I

L(l) -26 ll2) = -24

F -24 THERFCRE ~EIGHTS CN E.P. (3 1 0) OF DIVISION 2 ENTERS bASI~

STEP 3. STUPPING RULE.

F = -24

68

.t·,

PAGE 25

STEP 4. 1Al2) * E .P.I 12 41 IJI I 61

A' = 1---------1 = 16 41 * I Ol = 1181 I I I 0 I OJ

1 I 11

STEP 5. 110 -11 OJ I 61 I 61

t:H * A1 = lo 1 -13 Ol * 1181 = 1181 IO 0 1 Ol I Ol I Ol T~E KATlOS ARE : I o o 0 11 I 11 Ill

9/6 ' 12/18 ' J../0 ' J../1 11 0 -11 Ol 1201 I 91

81 * 8' = IO 1 -13 Ol * J25l = 1121 I o o 1 Ol I 11 I 11 JO o 0 11 I 11 I 11

THE MINIMUM RATIG IS 12/18. R = 2. CB = (0,24,26,0) THE EXTRI::ME PGII'oTS lh THE &A!l.IS ARi:: : (_,_) (3,0) (2,~) (_,_)

STEP t.: a INVEKSE 11 -1/3 -20/3 Ol

JO 1/18 -13/18 Ol IO 0 1 Ol IO -1/18 13/18 11

Pll = l 0,4/3)

69

PAGE 27

QUI: SHUN 15 : IF 11. -1./3 -20/3 OJ

B II\ VERSE = IO 1/l.d -13/lli Ol Pll '"' 10.4/3) IO 0 1 Ol IO -l./1d 13/18 11

ll 31 12 41 A( 1) = IC: 31 A(2) = lo 41 (.( u = (it 6) AND CCZ)

COMPUTE THE COEFFICIENTS UF THE CBJECTIVE FUNCTION FLH Ll~ISit~ 1 A~O DIVISIOh 2 RcSPELTlVELV.

A. (O,OJ AND (Q,Q) B. (-4,-6) ANU (-8,-!:i) C. (-~/3,-2) AND (0,1./3) D. (-16/3,-1) AND (4,-2/3)

PAGE Zd

*** lTtRATluN 2

STEP 1. UlVISlUN 1 CeJ. CCI:rFICli:NTS

(0,4/~) ..)< 1131- (4,6) = l-4/3,-i) 12 31

SuLUT I ON X(l) = 2 X(Z) = 3

DIVISION 2 C~J. COEFFICIENTS '0,4/3) * 12. 41 - (&,5)

16 41

SuLUTI ~ Y( U = 0 '(( 2) 0

z' l) = -2.6/3

ztz l = o

70

= (8 51

PAGE 29

STEP 2. 1-Z0/3 I

PIO: (0,24 1 26,0) * l-13/181 26/3 I 1 I I 13/Ull

l(1) + 26/3 = 0 I o I

PI o = I o , 2 4, 26 , o J * I o I o I o I Ill

Ll.:l+O 0

f = i)

S Tt'P 3. :; T G P t' I Mi K Ul E •

f IS.>= 0 TI-llS IS Afll CPT! MAL SC.lUTICN.

PAGE .:)()

THt eXTREME PCI,.,TS II'.! THE nASIS ARE : (_,_) (_j,Q) 12,3) (_,_) Trlt wtiGHTS CN THESE PCINTS AkE : 5 , 213 , 1 , l/3

CLMFuTED BY E INVE~SE * CkiGINAl RHS

X= SUM LIJ) * XIJ) : 1 * 12,3) 12,3) = Xll),X(2J

Y ~UM L(JJ * Y(J) = 213 * IJ,OJ = 12,0) = YllJ,Y(2)

fHUS, AI'~ UPTIMAl SClUTICN FCK THIS PR(BlEM IS X(l) 2 , X(2) = 3 Ylll .: , Y(2) = 0

l 4*2 + 6*3 + 8*2 + 5*0 = 42

71

APPENDIX F

LISTING OF THE CONTROL PROGRAM

72

0001 0002 0003 0~

0005 0006 0001 oooa 0009 0010 0011 0012 0013 0014 OOlS 0016 C017 0018 0019 0020 0021 0022 0023 0024 C025 0026 0027 0028 0029 0030 OC31 0032 0033 CC34 0035 0036 CC37 0038 0039 C040 C04l OOitZ C0Jt3 0044 0045 0046 001t7 CC48 0049 005Q C051 CC52 0053 0054 C055 0056 0057 C058 C0 59 0060

I* GENERAL INSTRUCTIOa5 *I

C QHROL PRO,.T MAIN WRITE THIS PROGRAM IS DESIGNED TO OPERATE ON ANY TSO TERMINAL. WRITE IT IS INTEAACTIVE. MEANING THE USER WILL BE PROMPTED FOR A WRITE RESPONSE. A OECWRITER IS PREFERRED SINCE YOU CAN HAl~TAlN WRITE A HARDCOPY OF THE SESSION 4ND REFER TO IT AT ANY TIME. WRITE HOWEVER, DECSCOPES AND IBM 3277 1 5 CAN ALSO BE USEe. WRITE THE OPERATION OF A OECSCUPE AND DECWRITER IS SLIGHTLY WRITE DIFFERENT THAN A 3277. IF YOU ARE USING A CECSCOPE OR WRITE DECWRITER. AFTER TYPIN& A RESPONSE PRESS THE RETURN KEY. WRITE I'IJWEVER, WITH THE 32.71 YOU HUST CLEAR THE SCREEN fiRST WRITE THEN ENTER YOUR RESPONSE. THE INSTRUCTIONS DURIN- A WRITE SESSION ASSUME YOU ARE USING AN 18M 3277.

I* BEGINNING CF PROGRA.. *I

PARTU + WRITE IF YOU ARE USIN5 A 3277 OR ANYTHIN5 SIMILAR ENTER - CRT READ &TERM IF & TERM ~ CRT TMEN + DO

WRITENR PRESS CLEAR AND ENTER READ &REPLY

END L 1 TSO.Ul6300A.PAGE.DATA• 10 190 SHU .. READ &ANS DO WHILE U.ANS _. YES) AND lUNS ._ NO)

~RITE INVALID ANSWER &ANS - REENTER READ &ANS

EfoiD lf £.TERM a CRT THEN WRI TENR PRESS CLEAR THEN WRITE TYPE IN YOUR NAHE AND HIT ENTER.

I* READ STUDENT'S HAM£ •1

READ &NAME IF &ANS "' NO THEN GOTO PART2

I* INTRODUCTION *I

L 1 TSO.Ul6300A.PAGE.DATA• 250 430 SHUll READ UNS LBL3: + l 1 TSO.Ul6300A.PAGE.DATA 1 470 650 SHU .. READ &ANS IF l&ANS a STDPI THEN &OTO L8Ll

I* ANGULAA STRUCTURE *I

L 1 TSO.Ul6300A.PAGE.DATA• 690 880 SNU .. READ &ANS

I* PRINT QUESTION 11 *I

L 'TSO.Ul6300A.PAGf.DATA• 990 1050 ~ WRITENR ANS • READ UNS SET &CNT • 2 DO WHILE l&CHT > 01 AND ClAN$,_ MULTIDIVISJOMA4)

73

0061 0062 0063 0064 0065 C066 0067 0068 OC69 0070 0071 0012 0073 OOllt 0075 CC76 0071 0078 0079 0080 0081 0082 0083 CC81t 0085 0086 0087 0088 0089 0090 0()q1 C09Z Co<33 0094 C095 C096 Q()ql 0098 0099 0100 0101 0102 0103 0104 0105 010. 0107 0108 0109 oua 0111 Oll2 0113 Cll4 Ol15 0116 0117 Cll8 Cll9 0120

WRITE liRONGe TRY AGAIN &NAME. ~ITE AND WATCH FOR SPELLIN- OR TRY A SIMILAR WOAD WR I TEHR ANS • READ UNS SET "NT • 'CNT - 1

END

I* GI~! A CHANCE TO REREAD PREVIOUS PA'E

IF &ANS ~ "ULTIDIVISIONAL THEN WRITE VERY GOOD 'NAME IF &CNT • 0 THEN + DO

WRITE WOULD YOU LIKE TO REREAD THE PREVIOUS PAGE LNAME1 READ &ANS IF &ANS • YES THEN + 00

IF &TERK ,_ CRT THEN GOTO LBL3 WRITENR PRESS CLEAR AND ENTER. READ &REPLY GOTO L6L3

END END IF &TERM • CRT THEN + 00

kRITENR PRESS CLEAR AND HIJ ENTER. READ &REPLY

EhD

1• ANSWER TO QUESTION Jl *I

L 1 TSO.U16300A.PAGC.DATA' 1140 1320 SNU8 READ &ANS IF l&ANS a STOP) THEN GOTO LBL1

1• INTRODUCTION TO REFORMULATIOA *I

SET &CNT • 2 L 1 TSO.U16300A.PA&E.DATA• 131t0 1500 SHU8 READ &ANS If l'ANS • STOPl THEN GOTO LBLI

PRINT QUESTION J2

L 'TSO.U16300A.PA&E.DATA• 1580 1130 SHUa WRITEHR ANS a READ &ANS 00 WHILE l&ANS ,. BJ AND l&CNT > OJ

.,

WRITE SORRY &NAME. &ANS IS AN INCORRECT ANSWEa. WRITE TRY AGAIN. YOU HAVE &CNT MORE CWANCES. If &TERM a CRT THEN + 00

WRITENR PRESS CLEAR AND HIT ENTER. READ f'Jl.EPL Y L 1 TSO.Ul6300A.PAGE.OATA' 1580 1730 SNU"

END WRIT E!lll ANS a

SET &CNT • &CNT - 1 READ UNS

END

74

.,

75

0121 I* ANSWER TO QUESTION ta *I 0122 0123 IF &ANS • 8 THEN WRITE VERY GOOD &NAME. 012~ WRITE THE CORRECT ANSWER IS 8. 0125 WRITE YOU WOULD EVALUATE THE VIOLATIONS ' CALtuUTE PENALTIES.-0126 WRITE BUT HOW? 0127 IF &TERM • CRT THEN WRITENR PRESS CLEAR AND 0128 WRITENR HIT ENTER. 0129 READ UNS 0130 If l&ANS • SlOPa THEN GQTO LBLl 0131 0132 I* DEFINE A GENERAL PROBLEM *I 0133 0134 L 'TSO.U16300A.PAGE.DATA' 1170 1970 SHUM 0135 WRITE IF YOU ~OULD LIKE TO SEE THE PROBLEM REFORMULATED 0136 WRITE BY USING THESE CONSTRAINTS ENTER YES. 0137 READ &ANS 0138 IF &TERM a CRT THEM + 0139 DO Olo\0 WRITENR PRESS CLE.AA AND HIT ENTEA. 0141 READ &REPLY 0142 END 0143 Ollt4 I* REFORMUL.AT II:* 1M MOAE DETAIL *I 0145 0146 IF l&ANS :a YeSJ THEN + 0147 DO OH8 L 'TSO. U16300A. PAGE.DATA1 2000 2200 SIIIUM 0149 READ &ANS 0150 ENO 0151 IF l&ANS • STOP) THEN GOTO LBLl 0152 0153 I* PRINT QUESTION 13 *I 0154 0155 L 1 TSO.U16300A.PAGE.DATA' 2290 2400 SHUll 0156 WRITENR ANS • 0157 READ &ANS 0158 SET &CNT a 2 0159 DO WHILE ( &CNT > oa AND ( UNS ._ D) 0160 WRITE SORRY &NAME, &ANS IS AN INCORRECT ANSWE& 0161 WRITE TRY AGAIN, YOU HAVE &CNT MORE CHANCES. 0162 IF &TERM = CRT THEN + 0163 DO 0164 WRITENR PRESS CLEAR AND HIT ENTER. 0165 READ &REPLY 0166 L 1 TSO.Ul6300A.PAGE.OATA 1 2290 2400 SNUR 0167 END 0168 WRITENR ANS 3

0169 READ &ANS 0170 SET &CNT a &CNT - 1 Ol1I END Cl72 0173 I* ANSWER TO QUESTION 13 *I 0174 0175 IF &ANS • D THEN WRITE VERY GOOD &NAME 0176 WRITE THE CORRECT ANSWE& IS D Cl77 IF l'TERM z CRT) THEN+ 0178 DO 0179 WRITENR PRESS CLEAR AND HIT ENTEa. 0180 READ &REPLY

0181 END 0182 0183 I* GIVE A CHANCE TO REREAD TMI PREVIOUS PA&I *I 018~ 0185 IF l&CNT • 0) AND l&ANS ,. Dl THEN + 0186 DO 0187 WRITE ~OULD YOU LIKE TO REREAD THE PREVIOUS PA&E7 0188 READ &ANS 0189 IF &TERM • CRT THEN WRITENR PRESS CLEAR AND 0190 WRITENR HIT ENTER. 0191 READ &REPLY 0192 IF &ANS = YES THEN L 1 TSO.Ul6300A.PAGE.OATA 1 2000 2208 SHUA 0193 READ &REPLY 019~ END 0195 0196 I* PARTITIONIN& Of SIMPLEX MULTIPLIERS *I 0197 0198 L 1 TSO.U16300A.PAGE.DATA' 2~80 2660 SNUA 0199 WRITENR DO YOU WANT TO LEARN HOM THIS IS DONE I• MORE DETAIL7 0200 READ &ANS 0201 0202 I* PARTITIONING N MORE DETAIL *I 0203 020~ IF l&ANS • YESJ THEN + C205 DO 0206 L 1 TSO.U16300A.PAGE.DATA 1 2670 2860 SHUM 0207 WRITENR PRESS ENTER TO CONTINUE OR TYPE STOP TQ TERMINATE. 0208 READ &ANS 0209 END 0210 IF l&ANS s STOPI THEN GOTO LBLl 0211 0212 I* PRINT QUESTION I~ *I 0213 C214 L 1 TSO.Ul6300A.PAGE.DATA 1 2930 30~0 SNUA 0215 WRITENR ANS ~ 0216 READ &ANS 0217 SET &CNT z 2 0218 DO WHILE l&CNT > OJ AND CLANS ,. A) 0219 WRITE SORRY &NAME, &ANS IS AN INCORRE~T ANSWEI. 0220 WRITE TRY AGAIN, YOU HAVE &CHT MORE C~ANCES. 0221 IF &TERM • CRT THEN + 0222 DO 0223 WRITENR PRESS CLEAR AND ENTER. 022~ READ UEPL Y 0225 L 1 TSO.Ul6300A.PA6f.DATA 1 2930 3~0 SHUA 0226 END 0227 WRITENR ANS • 0228 READ &ANS C229 SET &CNT ~ &CNT - 1 0230 END 023J: 0232 I* ANSWER TO QUfST10N 14 •I 0233 023~ IF &AHS a A THEN WRITE VERY GOOD &NAME 0235 WRITE THE CORRECT ANSWER IS A. 0236 WRITE 8 IS AN ~*M MATRIX• 0231 ~RITE BUT TO CALCULATE Pil AND PIO YOU ME£D ONLW C238 WRITE NLC+1 COLI.JIINS OJ= B. 0239 0240 I* GIVE A CHANCE TO REREAD THE PREVIOUS PA5E *I

02~1 0242 0243 024~ C245 0246 0247 0248 0249 0250 C251 0252 0253 0254 0255 0256 0257 0258 0259 C260 0261 0262 0263 0264 0265 0266 0267 C268 0269 0270 0271 0272 0273 0274 C275 C21b 0277 0278 0279 0280 0281 0.282 C283 0284 0285 0286 0287 0288 0289 0290 C29l 0.292 0293 C294 0295 0296 0297 0.298 0.299 0300

IF C&CHT • Ol AND C&ANS,. AI THEN+ DO

~RITE WOULD YOU LIKE TO REREAD THE PREVIOUS P&GE1 READ &AMS IF CUNS • YESI THEN + DO

IF &TERM a CRT THEN WRITENR PRESS CLEAR AND WRITENR HIT ENTER. READ &REPLY L 1 TSO.U16300A.PAGE.DATA• 2670 2870 SHUN READ &REPLY

END END WRITENR WOULD YQJ LIKE TQ SEE A SIMPLE ALGORITHM AND EXAI4PLf1 READ UNS IF &TERM z CRT THEN + DO

WRITENR PRESS ClEAR AND Hll ENTER. ~EAD &REPLY

END IF l&ANS ,. YESI THEN GOTD LBLl

I* BEGINNING OF A~GORITWM *I

L 1 TSO.Ul6300A.PAGE.OATA 1 3110 3300 SNUR READ &ANS L 1 TSO.Ul6300A.PAGE.OATA• 3330 3520 SNUM WRITENR WOULD YOU LIKE TO SEE AH EXAMPLE OF THIS AlGORITHM? READ &ANS IF &ANS ~ NO THEN GOTO l8Ll

,. PRINT THE EXAMPLE ., LBL2: + L 1 TSO.Ul6300A.PAGE.OATA• 3550 3740 SNUR WRITENR PRESS ENTER. READ &REPLY L 'TSO.U16300A.PAGE.DATA• 3810 3940 SNUB

INITIALIZE ., IF &TERM • CRT THEN WRITENR PRESS CLEAR AND WRlTENR HIT ENTER. READ &REPU

,. ,. ** ITEKA TION 0 STEP 1

L 1 TSO.Ul6300A.PAGf.OATA• 3990 4040 SNUR

., ., WRITE~ DO YOU WANT TO SEE HOW THE SOLUTION IS t0MPUTED7 RfAO &ANS IF &TERM a C~T THEN + DO

wRITENR PRESS CLEAR AND HIT ENTER. EN~EAO &REPL'

I* STEP 1 IN MOaE DETAIL *'

0301 0302 0303 0304 0305 0306 0307 0308 0309 0310 0311 0312 0313 0314 0315 0316 0311 0318 0319 0320 0321 0322 0323 0324 0325 0326 0327 0328 0329 0330 0331 0332 0333 0334 0335 0336 0337 0338 0339 0340 0341 0342 u343 0344 0345 0346 0347 C348 0349 0350 0351 0352 0353 0354 0355 0356 0357 0358 0359 0360

If &ANS • YES THEN + DO

L 1 TSO.U16300A.PAGE.DATA• 4050 4170 SMUR READ &ANS

END

,. STEP 2 ., L 1 TSO.Ul6300A.PlGE.OATA• 4220 4280 SNU8 WRITENR 00 YOU NEED HELP2 READ UNS

1• STEP 2 IN MORE DETAIL *I

IF C&ANS • YES) THEN L 1 TSO.U1b300A.PAGE.DATA• 4290 4330 SNUM

1• STEP 3 */

L 'TSO.U16300A.PAGE.DATA' 4330 4360 St«JII If &TERM a CRT THEN WRITENR PRESS CLEA~ AND WRITENR HIT ENTE~. READ &REPLY

1• STEP 4 *I

L 1 TSO.U16300A.PAGE.OATA 1 4430 ~60 SNUA WRITENR IF YOU HEED HELP TO GENERATE THi COLUMN ENTER YES. READ &ANS

,. STEP 4 IN MORE DETAIL ., IF l&ANS z YESJ THEN L 1 TSO.Ul6300A.PAGE.OATA• 4470 4500 SNUM IF &TERM = CRT THEN WRlTENR PRESS CLEAR AND WRITENR HIT ENTER. READ &REPLY

,. STEP 5 ., L 1 TSO.U16300A.PAGE.DATA• 4520 4620 SNUa IF &TERM a CRT THEN WRlTENR PRESS CLEAR AND WRITENR HIT ENTER. READ GREPLY

STEP 6 ., L 'TSO.Ul6300A.PAGE.OATA• 4660 4700 St«JII WRITE THERE ARE ~ANY WAYS TO FINO AN INVERSE. WRITE~ WOULD YOU LIKE TO SEE AN EASY O~E1 READ UNS

I* EASY WAY TO FIND AN 1NYER5a *I

If l&ANS • YES) THEN L 1 TSO.Ul6300A.PA~E.DATA• 4110 4820 SHUM IF &TERM • CRT THEN WRITENR PRESS CLEAR AND WRITENR HIT ENTER. READ &REPLY

1• •• ITERATIOM 1 *I 1• STEP 1 *I

78

0361 0362 0363 036<\ 0365 0366 0367 0368 C369 0370 C31l 0372 0373 0374 C375 0376 0377 0378 C379 0380 0381 0382 C383 C381t 0385 0386 0387 0388 0389 C390 C391 03c;l C393 0394 0395 C396 0397 0398 0399 C400 0401 C402 Olt03 C404 C405 Clt06 Olt07 C408 0409 0410 0411 0412 0413 Oltllt 0415 Olt16 Olt17 0418 0419 Olt20

l 1 TSO.Ul6300A.PAGE.DATA• lt890 lt930 SNUa WRITE~ DO YOU NEED f'IORE INFORMATIOd READ &ANS

I* STEP 1 IN MORi DEJA~ *I

IF I&ANS • YESl THEN L 1 TSO.Ul6300A.PAGE.DATA1 lt950 5020 SNUM IF &TERM • CRT THEN WRITENR PRESS CLEAR AND WRITENR HIT ENTER. READ &REPLY

I* STEPS 2e3 *I

L 1 TSO.U16300A.PAGE.OATA 1 5090 5280 SHU. WRITENR DO WE STOP CR CONTINUEW READ &ANS IF C&ANS = STOP) THEN WRITE NO. -24 < O. WE MUST CONTINUE. IF &TERM = CRT THEN WRlTENR PRESS CLEAR AND WRITENR HIT ENTER. READ &REPLY

STEPS 'te5 ., l 1 TSO.Ul6300A.PAGE.OATA• 5300 5510 SNU8 READ &REPLY

I* STEP 6 *I

L 1 TSO.Ul6300A.PAGE.DATA• 5540 5600 SNUM

79

WRITE WORK THIS YOURSELF TO BE SURE YOU KNOW HOW IT 1 S CONE. wRITE 00 't'OU WANT TO START OVER AT THE BEGINNIN~ Of THE EXAMPLf1 READ &ANS

AT THIS POINT YOU CAN START OVER

IF I&ANS = YESJ THEN + DO

IF &TERM • CRT THEN + DO

WRITENR PRESS CLEAR AND HIT ENTER. READ &REPLY

END GOTO LBL2

EIIID IF &TERM • CRT THEN + DO

wRITE~R PRESS CLEAR AND HIT ENTER. READ &REPLY

END

,. PRINT QUESTION 15

L 1 TSO.Ul6300A.PA,E.DATA• 5740 5900 ~· WRITENR ANS • READ &ANS SET &CNT • 2 00 WHILE I&CNT > OJ AND l~NS ,. Cl

.,

WRITE:SORRY &NAME• &ANS IS AN INCORRECT ANSWE&. wRITE TRY AGAIN• YOU HAVE 'CNT MORE CNANCES. IF &TERM • CRT THEN +

.,

0<\21 0<\22 0423 0424 0425 0426 0427 0428 0429 0430 0431 0432 0433 0434 0435 0436 0437 0438 0439 0440 0441 C442 04't3

0"'" C445 0446 Olt47 0448 C449 C450 0451 0452 0453 C454 G455 G456 0457 0458 0459 0460 G461 C462 0463 C46tt 0465 0466 0467 0468 0469 0470 0471 0472 0413 C474 0475 C476 C471 0478 0479 0480

00 WRITENR PRESS CLEAR AND HIT ENTEA. READ &REPLY L 1 TSO.U16300A.PAGE.DATA 1 5740 5900 SNUM

END WRI TENR ANS • READ UNS SET &CNT • I.CNT - 1

END

I* ANSWER TO QUESTION 15 *I

IF &ANS 2 t THEN WRITE VERY GOOD 'NAME. WRITE THE CORRECT ANSWER IS C. WRITE PI1 *Alii - Clll

I* GIVE A CHANCE TO REREAD THE PREVIOUS PAGi *I

IF &CNT a 0 THEN + DO

WRITE DO YOU THINK YOU SHOULD START OVER AI THE BEGINNING! READ &ANS IF &ANS ~ YES THEN GOTO PARTl

80

WRITE DON'T YOU THINK YOU SHOULD AT LEAST REREAD THE EXAMPLE? READ &ANS IF £.ANS = YES THEN GOTO LBL2 WRITE OK• LET 1 S CONTINUE. ~·RE ALMOST DONE.

Er.o IF &TERM = CRT THEN WRITEkR PRESS ClEAR AND WRITENR HIT ENTER. READ &REPLY

I* ** ITERATION 2 *I I* STEP 1 *I

L 1 TSO.U16300A.PAGE.DATA 1 5980 6140 SNUa If &TERM = CRT THEN WRITENR PRESS ClEAR AHD WRITENR HIT ENTER. READ &REPLY

STEPS 2.3 OPTIMAL SOLUTION

L 1 TSO.Ul6300A. PAGE.DATA1 6180 6370 SNUa IF &TERM z CRT THEN WRITENR PRESS CLEAR AND WRITENR HIT ENTER. READ &REPLY

,. COMPUTE THE AIISWEa

L 1 TSO.Ul6300A.PAGE.DATA 1 6400 6580 SHU. IF &TERM ~ CRT THEN WRITENR PRESS CLEAR AND WRITENR HIT ENTER. READ &REPLY

I* END r:# EX.UPL& AND PAAT1 *I

LBLl: + WRITE IF YOU HAYE DATA YOU WANT TO RUN AS A PROGaAA ENTEA YES. READ UNS IF (&ANS ~ YES) THE» GOTG FIN

0~81 01Ht2 O"ta3 0481t Olt85 0486 0481 0~88 0489 0490 0491 C492 C493 0491t 0495 Clt96 0497 0-498 C499 C500 0501 C502 0503 C501t C505 0506 C507 C508 C509 0510 0511 C512 0513 0514 0515 C516 0~17 0518 0519 0520 0521 C522 C523 C521t 0525 0526 C527 0~28

C529 0530 0531' 0532 0533 0531t C535 0536 0537 0538 0539 0540

, .................... . P~RT21 + WRITE ENTER A TITLE TO YOUR PROSLEM READ UITLE FREE FILEIFT05FOOl,FT06FOOl) ALLOC DAl*) FllFT05FOOl) SHR Allot OAl*) FilFT06f001) SHR LCAOGO JESSE.OBJ FORTLIB WRITE &NAME, IS THERE ANOTHER PR06LEM YOU WANT TO RUN? READ &ANS FREE FILEifT05FOOl,FT06F001) IF I&ANS = YES) THEN GOTO PART2 WRITE DO YOU WANT TO GO THROUGH PAA.Tl AGAIN? READ &ANS If I&ANS = YES) THEN GOTO PARTl FIN: + WRITE TO END THE SESSION ENTER LOGOFF END

8I

VITA - ,2_.

William Arthur Senters

Candidate for the Degree of

Master of Science

Thesis: A TSO PRESENTATION OF A DECOMPOSITION TECHNIQUE FOR SOLVING LARGE-SCALE MULTIDIVISIONAL LINEAR PROGRAMMING PROBLEMS

Major Field: Computing and Information Sciences

Biographical:

Personal Data: Born in Midwest City, Oklahoma, July 8, 1948, the son of Mr. and Mrs. Charles D. Senters.

Education: Graduated from Collegia del Espiritu Santo, San Juan, Puerto Rico, in May, 1966; attended Eastern New Mexico University, Portales, New Mexico, from Sept., 1970 to May, 1972; attended Chapman College, Orange, California, from Jan., 1973 to Dec., 1973; received Bachelor of Science degree from Oklahoma State University at Stillwater, Oklahoma, in May, 1975, with a major in Business Administration; completed requirements for a Master of Science degree at Oklahoma State University in May, 1978.

Professional Experience: Member of the United States Air Force from Dec., 1969 to Dec., 1973; graduate teacher of 'Introduction to Data Processing' in the College of Business at Oklahoma State University, Stillw~ter, Oklahoma from Sept., 1976 to May, 1977. Applications programmer, Texaco, Inc., Houston, Texas, January, 1978 to the present.

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